Class 10th Science Gujarat Board Solution
Exercise- What is the wavelength range of visible light?A. 4 x 10 - 7m to 8x10 - 7 m B. 4x10 - 9 m…
- What is the relation between radius of curvature (R) and the focal length (f) of a…
- Which type of reflection will be represented by a light reflected from a book?A. Regular…
- Through which of the following points, a ray passing through a centre of curvature and…
- At what distance in front of the concave mirror should an object be placed to get its…
- The magnification of plane mirror is alwaysA. More than 1 B. 1 C. Less than 1 D. Zero…
- The focal length of plane mirror is......A. Zero B. Infinity C. Uncertain D. Equal to…
- The distance between the object at 2 m from a plane mirror and its image is....A. 4 m B. 1…
- At what distance should an object be placed to obtain its real, inverted and of same…
- Which of the following materials has maximum optical density?A. Glass B. Water C. Pearl D.…
- The absolute refractive index of any medium is always...A. 1 B. 1 C. 1 D. Zero…
- Which of the lenses with focal length 10cm, 20cm, 25cm and 50cm has maximum power?A. 50 cm…
- What is the focal length of a convex lens having power + 5.0D?A. - 10cm B. - 20cm C. +…
- If the absolute refractive indices of water, benzene, and sapphire are 1.33, 1.50 and 1.77…
- Which type of an image is formed by a plane mirror?A. Real and inverted B. Real and erect…
- If the absolute refractive indices of water and glass are 4/3 and 3/2 respectively, then…
- The absolute refractive indices of water glass and diamond are 1.77, 1.50 and 2.42…
- Which of the following always form virtual image?A. Concave mirror and convex lens B.…
- What will be the angle of refraction for the light ray incident normal at the surface?A.…
- The compound microscope consists of two convex lenses of 5 cm and 20 cm focal length, and…
- What is called regular and irregular reflection of light?
- Write the laws of reflection of light.
- What are called centre of curvature and radius of curvature of mirror?…
- Draw a ray diagram showing position, nature and size of an image formed by concave mirror…
- Draw a ray - diagram showing position, nature and size of an image formed by concave…
- Draw a ray - diagram showing position, nature and size of an image formed by convex mirror…
- Obtain the position, nature and size of an image formed by a plane mirror from the formula…
- Write laws of refraction of light.
- What is called the absolute refractive index of a medium? Obtain the general from of…
- Draw a ray - diagram showing the position, nature and size of an image formed by convex…
- Draw a ray - diagram showing the position, nature and size of an image formed by convex…
- Draw a ray - diagram showing the position, nature and size of an image formed by concave…
- What is called the magnification of an image? Derive the formula of magnification for…
- Explain the reflection by a plane mirror by drawing suitable figure.…
- Give the Cartesian sign convention for the reflection by spherical mirror.…
- With the necessary figure, explain the refraction of light through a rectangular glass…
- Obtain the lens formula for spherical lens.
- Explain how the position of images is located for spherical mirror by considering the…
- Write a note on power of lens.
- Derive the formula for spherical mirror
- Explain the construction and working of a compound microscope with a neat ray - diagram.…
- Write a note on astronomical telescope.
- An object of height 5 cm is placed of 10cm from convex mirror of focal length 15cm.Find…
- An object of height 6 cm is placed at a distance of 15cm from concave mirror of focal…
- The rays of light are entering from glass into glycerine. If the absolute refractive…
- The refractive index of light entering glass from water is 1.12. Find absolute refractive…
- When the light entering from glass to water, refractive index of water with respect to…
- An object on place perpendicular to the principle axis of a convex lens having focal…
- An object is placed perpendicular to the principle axis of concave lens of focal length…
- A power of convex lens is + 4.0 D. At what distance should the object from the lens be…
- What is the wavelength range of visible light?A. 4 x 10 - 7m to 8x10 - 7 m B. 4x10 - 9 m…
- What is the relation between radius of curvature (R) and the focal length (f) of a…
- Which type of reflection will be represented by a light reflected from a book?A. Regular…
- Through which of the following points, a ray passing through a centre of curvature and…
- At what distance in front of the concave mirror should an object be placed to get its…
- The magnification of plane mirror is alwaysA. More than 1 B. 1 C. Less than 1 D. Zero…
- The focal length of plane mirror is......A. Zero B. Infinity C. Uncertain D. Equal to…
- The distance between the object at 2 m from a plane mirror and its image is....A. 4 m B. 1…
- At what distance should an object be placed to obtain its real, inverted and of same…
- Which of the following materials has maximum optical density?A. Glass B. Water C. Pearl D.…
- The absolute refractive index of any medium is always...A. 1 B. 1 C. 1 D. Zero…
- Which of the lenses with focal length 10cm, 20cm, 25cm and 50cm has maximum power?A. 50 cm…
- What is the focal length of a convex lens having power + 5.0D?A. - 10cm B. - 20cm C. +…
- If the absolute refractive indices of water, benzene, and sapphire are 1.33, 1.50 and 1.77…
- Which type of an image is formed by a plane mirror?A. Real and inverted B. Real and erect…
- If the absolute refractive indices of water and glass are 4/3 and 3/2 respectively, then…
- The absolute refractive indices of water glass and diamond are 1.77, 1.50 and 2.42…
- Which of the following always form virtual image?A. Concave mirror and convex lens B.…
- What will be the angle of refraction for the light ray incident normal at the surface?A.…
- The compound microscope consists of two convex lenses of 5 cm and 20 cm focal length, and…
- What is called regular and irregular reflection of light?
- Write the laws of reflection of light.
- What are called centre of curvature and radius of curvature of mirror?…
- Draw a ray diagram showing position, nature and size of an image formed by concave mirror…
- Draw a ray - diagram showing position, nature and size of an image formed by concave…
- Draw a ray - diagram showing position, nature and size of an image formed by convex mirror…
- Obtain the position, nature and size of an image formed by a plane mirror from the formula…
- Write laws of refraction of light.
- What is called the absolute refractive index of a medium? Obtain the general from of…
- Draw a ray - diagram showing the position, nature and size of an image formed by convex…
- Draw a ray - diagram showing the position, nature and size of an image formed by convex…
- Draw a ray - diagram showing the position, nature and size of an image formed by concave…
- What is called the magnification of an image? Derive the formula of magnification for…
- Explain the reflection by a plane mirror by drawing suitable figure.…
- Give the Cartesian sign convention for the reflection by spherical mirror.…
- With the necessary figure, explain the refraction of light through a rectangular glass…
- Obtain the lens formula for spherical lens.
- Explain how the position of images is located for spherical mirror by considering the…
- Write a note on power of lens.
- Derive the formula for spherical mirror
- Explain the construction and working of a compound microscope with a neat ray - diagram.…
- Write a note on astronomical telescope.
- An object of height 5 cm is placed of 10cm from convex mirror of focal length 15cm.Find…
- An object of height 6 cm is placed at a distance of 15cm from concave mirror of focal…
- The rays of light are entering from glass into glycerine. If the absolute refractive…
- The refractive index of light entering glass from water is 1.12. Find absolute refractive…
- When the light entering from glass to water, refractive index of water with respect to…
- An object on place perpendicular to the principle axis of a convex lens having focal…
- An object is placed perpendicular to the principle axis of concave lens of focal length…
- A power of convex lens is + 4.0 D. At what distance should the object from the lens be…
Exercise
Question 1.What is the wavelength range of visible light?
A. 4 x 10 - 7m to 8x10 - 7 m
B. 4x10 - 9 m to 8x10 - 9 m
C. 4x10 - 5 m to 8x10 - 5 m
D. 4x10 - 6 m to 8x10 - 6 m
Answer:Option (A): it is a fact that Wavelength of visible light is between 4 × 10 - 7 and 8 × 10 - 7. Therefore, Option A is correct.
Question 2.What is the relation between radius of curvature (R) and the focal length (f) of a spherical mirror?
A. R = f/2
B. R = f
C. R = 2f
D. R = 3f
Answer:Radius of curvature (R) is equal to twice the focal length that means R = 2f. Therefore, Option C is correct.
Question 3.Which type of reflection will be represented by a light reflected from a book?
A. Regular
B. Irregular
C. Both types
D. None
Answer:Surface of Book is rough, therefore the reflection from book is irregular because the reflected light beam will not go in the particular direction.
Since book has rough surface therefore Option B is correct.
Question 4.Through which of the following points, a ray passing through a centre of curvature and reflected by concave mirror will pass through?
A. Focus
B. Centre of curvature
C. Pole
D. All
Answer:When ray of light passes through the centre of curvature of the concave mirror it strikes along the normal that means it is incident on mirror at 900.Hence the incident ray coincides with the normal and retraces its path i.e. passes through centre of curvature.
As other options don’t go by above fact, so they are incorrect.
Question 5.At what distance in front of the concave mirror should an object be placed to get its virtual and erect image?
A. At centre of curvature
B. Beyond centre of curvature
C. Between focus and pole
D. At focus
Answer:when an object is placed between the focus and pole of concave mirror the image is formed behind the concave mirror and is virtual, erect and enlarged. As shown in figure below.
Option (A): When object is kept at centre of curvature than image is formed at centre of curvature and is real, inverted. Therefore, Option A is incorrect.
Option (B): when an object is kept beyond centre of curvature the image formed is real, inverted and diminished. Therefore, Option B is incorrect.
Option (C): when an object is placed between focus and pole than image is virtual and erect as explained in explanation. Therefore, Option C is correct.
Option (D): when an object is placed at focus than image is formed at infinity and is real. Therefore, Option D is incorrect.
Question 6.The magnification of plane mirror is always
A. More than 1
B. 1
C. Less than 1
D. Zero
Answer:Magnification = 
And plane mirror forms the image of same size as that of object. (Height of Image = Height of Object).
Therefore, Magnification is 1
Question 7.The focal length of plane mirror is......
A. Zero
B. Infinity
C. Uncertain
D. Equal to object distance.
Answer:A plane mirror can be considered as a spherical mirror of infinite radius of curvature. And Focal length is half of radius of curvature therefore it is infinity.
Question 8.The distance between the object at 2 m from a plane mirror and its image is....
A. 4 m
B. 1 m
C. 2m
D. 3m
Answer:In plane mirror Object distance from the mirror is equal to the image distance from the mirror.
Since Object distance from mirror = 2 m
⇒ Image distance from mirror = 2 m.
∴ Distance between image and object = 2 m + 2 m = 4 m.
Question 9.At what distance should an object be placed to obtain its real, inverted and of same height as the objet by a convex lens?
A. At focus
B. Between focus and centre of curvature
C. At centre of curvature
D. Between optical centre and focus.
Answer:when an object is kept at centre of curvature of convex lens same size, real and inverted image of an object is formed at centre of curvature as shown in figure below.
Option (A): when an object is kept at focus than image is formed at infinity and is magnified. Therefore, Option A is incorrect.
Option (B): when an object is kept between focus and centre of curvature than image is formed on other side of lens and is magnified (enlarged). Therefore, Option B is incorrect.
Option (C): when an object is kept at centre of curvature than real, inverted image is formed at centre of curvature as explained in explanation. Therefore, option C is correct.
Option (D): when an object is kept between optical centre and focus the image formed is virtual and erect. Therefore, Option D is incorrect.
Question 10.Which of the following materials has maximum optical density?
A. Glass
B. Water
C. Pearl
D. Diamond
Answer:optical density is defined as the degree to which the refractive medium refracts the transmitted ray of light. Therefore, Medium having highest refractive index will have highest optical density.
For refractive indices,
Water (1.33) < pearl (1.46) < glass (1.52) < diamond (2.42)
Question 11.The absolute refractive index of any medium is always...
A. 1
B. >1
C. <1
D. Zero
Answer:Absolute refractive index of medium is defined as 
, and speed of light in vacuum is highest than its speed in any medium. Therefore, n>1.
Question 12.Which of the lenses with focal length 10cm, 20cm, 25cm and 50cm has maximum power?
A. 50 cm
B. 25 cm
C. 20 cm
D. 10cm
Answer:Power of lens = 
.Therefore if focal length is less power is more and if focal length is more power is less.
Option (A): 50 cm is the highest focal length in given options; hence it will have lowest power. Therefore, Option A is incorrect.
Option (B): 25 cm is not the lowest focal length; hence it will not have highest power. Therefore, Option B is incorrect.
Option (C): 20 cm is not the lowest focal length; hence it will not have highest power. Therefore, Option C is incorrect.
Option (D): 10 cm is the lowest focal length in given options. Hence it will have highest power. Therefore, Option D is correct.
Question 13.What is the focal length of a convex lens having power + 5.0D?
A. - 10cm
B. - 20cm
C. + 10cm
D. + 20cm
Answer:Power of lens = .
⇒ Focal length = 
1m = 100 cm.
⇒ 0.2 m = 0.2 × 100 = 20 cm.
∴ Focal length = 20 cm. (Positive for convex lens.)
Option (A): since focal length is positive for convex lens. Therefore, Option A is incorrect.
Option (B): since focal length is positive for convex lens. Therefore, Option B is incorrect.
Option (C): focal length is 20 cm not 10 cm therefore option (C) is incorrect.
Option (D): as explained in explanation focal length = + 20 cm therefore option D is correct.
Question 14.If the absolute refractive indices of water, benzene, and sapphire are 1.33, 1.50 and 1.77 respectively, then which medium has maximum relative refractive index?
A. Sapphire relative to water
B. Sapphire relative to benzene
C. Benzene relative to water
D. Water relative to benzenes
Answer:The relative refractive index between a pair of media is the ratio of their absolute refractive indices.
Let n1 = absolute refractive index of water; n2 = absolute refractive index of benzene; n3 = absolute refractive index of sapphire.
Thus,
It can be seen that n31 is maximum and corresponds to refractive index of sapphire relative to water.
Question 15.Which type of an image is formed by a plane mirror?
A. Real and inverted
B. Real and erect
C. Virtual and erect
D. Virtual and inverted
Answer:Option (A): Plane mirror never forms real and inverted image, therefore option (A) is incorrect.
Option (B): the image formed by plane mirror is virtual and erect it is not real; therefore option (B) is incorrect.
Option (C): the plane mirror always form the virtual, erect image of same size as that of an object, therefore (C) is the correct answer.
Option (D): image formed is erect not inverted; therefore (D) is not the correct answer.
Question 16.If the absolute refractive indices of water and glass are 4/3 and 3/2 respectively, then what will be the ratio of velocity of light in water to that of glass?
A. 2
B. 8/9
C. 9/8
D. 1/2
Answer:We know that 
∴ 
× 
⇒ 
⇒ 
Question 17.The absolute refractive indices of water glass and diamond are 1.77, 1.50 and 2.42 respectively, which medium is most optically denser?
A. Water
B. Glass
C. Diamond
D. None
Answer:optical density is defined as the degree to which the refractive medium refracts the transmitted ray of light. Therefore, Medium having highest refractive index will have highest optical density.
Question 18.Which of the following always form virtual image?
A. Concave mirror and convex lens
B. Convex mirror and concave lens
C. Convex mirror and convex lens
D. Concave mirror and Concave lens
Answer:Option (A): Concave mirror and convex lens
Forms both virtual as well as real images, hence A is not the correct answer.
Option (B): for all positions of object convex mirror and concave lens always forms virtual and erect images behind the mirror/lens, therefore B is the correct answer.
Option (C): convex lens not forms only virtual images it also forms real images, hence C is not the correct answer.
Option (D): Concave mirror forms both real and virtual images, hence D is not the correct answer.
Question 19.What will be the angle of refraction for the light ray incident normal at the surface?
A. 900
B. 600
C. 300
D. 00
Answer:since the light ray is incident normally to the surface, therefore it will pass without any refraction. Hence angle of refraction = 00.
Question 20.The compound microscope consists of two convex lenses of 5 cm and 20 cm focal length, and then which of them will be object lens and eye piece?
A. Object lens with 20cm focal length and eye piece with 5 cm focal length.
B. Object lens with 5 cm focal length and eye piece with 20 cm focal length.
C. Both should have 20 cm focal length.
D. Both should have 5 cm focal length.
Answer:For the compound microscope focal length of an objective lens is smaller than eye piece lens.f0 = objective lens focal length, fe = eye piece focal length.
⇒ f0 < fe
Question 21.What is called regular and irregular reflection of light?
Answer:Regular reflection: when a parallel beam of light is incident on a plane or smooth surface, a beam remains parallel after reflection in a specific direction. Such reflection is called regular reflection. The figure below illustrates regular reflection.
Example: Reflection of light by plane mirror.
Irregular reflection: when a parallel beam of light is incident on rough or irregular surface like Book, stone etc., the beam does not remains parallel but spreads over wide region after reflection. Such reflection is called irregular reflection. The figure below illustrates irregular reflection.
Example: we can see book, table, and chair etc. due to irregular Reflection of light these objects.
Question 22.Write the laws of reflection of light.
Answer:the laws of reflection of light is as follows.
The angle of incidence is equal to the angle of reflection that means 
.The figure below illustrates it.
The incident ray, reflected ray, and normal to the mirror at the point of incidence all lie in the same plane.
The figure below illustrates the law more clearly.
Question 23.What are called centre of curvature and radius of curvature of mirror?
Answer:Centre of curvature: the centre of a spherical shell from which the mirror is made is called as the Centre of curvature of mirror. It is denoted by ‘C’.
Radius of curvature: the radius of a spherical shell from which the mirror is made is called as the radius of curvature of mirror. It is denoted by ‘R’.
The figure below shows the centre and radius of curvature of concave lens.
Question 24.Draw a ray diagram showing position, nature and size of an image formed by concave mirror when the object is placed beyond the centre of curvature.
Answer:Ray Diagram:
Position of Image: Between Centre of curvature (C) and focus (F).
Nature of image: real and inverted.
Size of image: diminished.
Question 25.Draw a ray - diagram showing position, nature and size of an image formed by concave mirror when the object is placed between pole and principle focus.
Answer:Ray Diagram:
Position of Image: behind the mirror
Nature of image: virtual and erect
Size of image: Magnified (enlarged).
Question 26.Draw a ray - diagram showing position, nature and size of an image formed by convex mirror when the object is placed between infinite distance and pole.
Answer:Ray Diagram:
Position of Image: between pole (P) and principal focus (F) behind the mirror.
Nature of image: virtual and erect.
Size of image: diminished.
Question 27.Obtain the position, nature and size of an image formed by a plane mirror from the formula of magnification.
Answer:In case of plane mirror the magnification is + 1;
Positive value of magnification indicates that image formed by plane mirror is erect.
Also, Magnification = 
⇒ Image height = Object height.
Therefore, Image formed by plane mirror is of the same size as of the object.
Magnification = 
⇒ v = - u;
This shows that image formed by plane mirror is at same distance as the object but behind the mirror means image is virtual.
Thus, Image formed by plane mirror is virtual, erect and same as size of the object.
Question 28.Write laws of refraction of light.
Answer:The refraction of light occurs according to the following laws which are as follows.
The incident ray, refracted ray, and the normal to the surface separating two media at the point of incidence all lie in the same plane.
The ratio of sine of angle of incidence and sine of angle of refraction remains constant (
= constant) subject to certain conditions. This law is called Snell’s law of refraction.
Question 29.What is called the absolute refractive index of a medium? Obtain the general from of Snell’s law in terms of refractive indices of two media?
Answer:The refractive index of medium with respect to the vacuum is called absolute refractive index.
Snell’s law in terms of refractive indices of two media: Let,
N1 = absolute refractive index of medium 1.
N2 = absolute refractive index of medium 2.
V1 = velocity of light in medium 1.
V2 = velocity of light in medium 2.
C = velocity of light in vacuum.
Then
N1 = 
and N2 = 
∴ 
⇒ N21 = 
………….. (1)
Now According to Snell’s law
N21 = 
…………….. (2)
From (1) and (2)
This is the general form of Snell’s Law.
Question 30.Draw a ray - diagram showing the position, nature and size of an image formed by convex lens when the object is placed at centre of curvature of lens.
Answer:Ray Diagram:
Position of Image: on other side of lens at centre of curvature.
Nature of image: real and inverted
Size of image: Same as size of object.
Question 31.Draw a ray - diagram showing the position, nature and size of an image formed by convex lens when the object is placed between its optical centre and focus.
Answer:Ray Diagram:
Position of Image: beyond the centre of curvature (2F) on the same side of lens where object is kept
Nature of image: virtual and erect
Size of image: Magnified (enlarged).
Question 32.Draw a ray - diagram showing the position, nature and size of an image formed by concave lens when the object is placed between at optical centre and infinite point.
Answer:Ray Diagram:
Position of Image: between focus (F) and optical centre (C) on the same side of lens where object is kept.
Nature of image: virtual and erect
Size of image: small (diminished).
Question 33.What is called the magnification of an image? Derive the formula of magnification for spherical lens.
Answer:Magnification of an image is defined as the ratio of the height of image to the height of object.
Magnification (m) = 
In the figure above by geometry we can easily see that ΔABO and ΔA’B’O’ are similar.
∴ 
…………… (1)
From definition of magnification.
Magnification (m) =
⇒ 
……… (2)
From (1) and (2)
Magnification = 
which is the formula for magnification of lens.
Question 34.Explain the reflection by a plane mirror by drawing suitable figure.
Answer:
MM’ = plane mirror.
AB = object of size ‘h’ on the left - hand side of the mirror at a distance ‘u’.
AP is the incident ray of light from point A of the object which falls on the mirror at point P. This incident ray is reflected back in the same path PA as shown in the figure above.
Another incident ray OC falls on the mirror at a point O and is reflected along the path OC.
Now, since reflected rays PA and OC are diverging and therefore cannot meet each other in front of the mirror, hence we extend these rays behind the mirror by dotted lines as shown in the figure below.
On extending these rays behind the mirror, we see that these rays meet at point A’ at a distance V’. Therefore A’ is the virtual image of point A of the object AB.
Similarly, a virtual image of point B will be formed behind the mirror as B’.
Now, to get a complete image of the object AB, we join the point A and B to point A’ and B’ by a dotted line.
We find that the image A’B’ being formed is virtual, erect and of same shape and size as the object AB; thereby giving us the characteristics of images formed by the plane mirror.
Question 35.Give the Cartesian sign convention for the reflection by spherical mirror.
Answer:Pole (P) of the mirror is taken as the origin and the Principal axis is taken as the horizontal X - axis. The sign conventions are as follows:
The object is always placed on the left side of the mirror.
All the distances measured to the right side of the origin are taken as positive, while the distances measured to the left side of the origin are taken as negative.
All the distances parallel to the principal axis of the mirror are measured from the pole of the mirror.
Distances measured vertically upwards from the principal axis of the mirror are taken as positive.
Distances measured vertically downwards from the principal axis of the mirror are taken as negative.
The focal length of Concave mirror is negative while that of convex mirror is positive.
The figure below illustrates the conventions used.
Question 36.With the necessary figure, explain the refraction of light through a rectangular glass slab.
Answer:Consider the rectangular slab as shown in the figure below.
AE = incident ray;
i = Angle if incidence;
r = Angle of refraction;
On entering the glass slab the light ray bends towards the normal and travels along EF.
The refracted ray EF is incident on the face SR at angle of incidence r’.
The emergent ray FD bends away from normal at angle of refraction e.
The emergent ray FD is parallel to the incident ray AE but it has been laterally displaced with respect to incident ray.
Question 37.Obtain the lens formula for spherical lens.
Answer:Consider the image formation by convex lens as shown in the figure below.
By Cartesian sign convention;
Object distance (OB) = - u;
Image distance (OB’) = + v;
Focal length (OF2 = OF2) = f
From the geometry we can see that in the figure above right triangle ΔABO and ΔA’B’O’ are similar.
∴ 
…… (1)
Also, in figure above by geometry right triangle ΔODF2 and ΔBAF2 are similar.
∴ 
∴ 
(∵ OD = AB, opposite sides of rectangle ABOD are equal).
∴ 
∴ 
………. (2)
From (1) and (2)
⇒ 
⇒ 
.
Dividing each term by uvf, we get.
⇒ 
⇒ 
This equation is called the ‘Lens Formula’
Question 38.Explain how the position of images is located for spherical mirror by considering the different rays using necessary ray - diagrams.
Answer:The image formed by spherical mirrors can be located by constructing ray diagrams. The point of intersection of two rays will give the position of the image.
Following rays can be taken into consideration while locating the image.
1. A ray parallel to the principal axis after reflection passes through the focus or appears to diverge from the focus. As shown in the figure below.
2. A ray passing through the principal focus of the concave mirror or ray which is directed towards the principal focus of convex mirror will emerge parallel to the principal axis as shown in the figure below.
3. A ray passing through the centre of curvature of the concave mirror or directed towards the centre of curvature of convex mirror after reflection is, reflected along the same path, as shown in the figure below.
4. A light ray incident obliquely to the principal axis towards the pole of a concave mirror or convex mirror is reflected obliquely following the laws of reflection. The figure below shows it.
Question 39.Write a note on power of lens.
Answer:Power of lens is defined as the efficiency with which a lens can converge or diverge the light ray.
The reciprocal of focal length of the lens is called the power of the lens (P).
P (in dioptre) = 
The S.I unit of power of lens is dioptre denoted by D.
Power of convex lens is taken as positive and of concave lens is negative. The instrument used for measuring the power of lens is called dioptremeter.
Question 40.Derive the formula for spherical mirror
Answer:
According to the Cartesian system in the above figure.
Object distance (PB) = - u
Image distance (PB’) = - v
Focal length (PF) = - f
Radius of curvature (PC) = - R
From the geometry we can see that in the figure above right triangle ΔABP and ΔA’B’P’ are similar.
∴ 
∴ 
……. (1)
Also ΔABC and ΔA’B’C’ are similar.
∴ 
……. (2)
Also
From above figure
CB’ = PC – PB’ = - R - (- v) = - R + v.
And
CB = PB – PC = - u - (- R) = - - - u + R.
From (2)
∴ 
⇒ R (u + v) = 2uv
Dividing both sides by Ruv we get.
We know that f = 
Putting this in the above equation we get
This equation is the ‘Mirror Formula’
Question 41.Explain the construction and working of a compound microscope with a neat ray - diagram.
Answer:Construction: A compound microscope consists of two convex lenses. The lens towards object is called objective lens and the lens near eye is called the eye piece. The focal length of objective lens is less than the focal length of eye piece. This combination forms the magnified image of the object.
The figure below shows the compound microscope.
Working: The object AB is placed in front of objective lens in such a way that its real, inverted and magnified image A’B’ is obtained beyond the centre of curvature of objective lens.
The image A’B’ becomes an object for the eye piece. The position of image is adjusted such that it lies within the focal length of eye piece.
The eye piece forms the virtual, erect and magnified image A”B” of the object.
Question 42.Write a note on astronomical telescope.
Answer:An astronomical telescope is an optical instrument which is used to see the magnified image of distant heavenly bodies like stars, planets, satellites and galaxies etc.
The final image formed by an astronomical telescope is always virtual, inverted and magnified.
Principle: An astronomical telescope works on the principle that when an object to be magnified is placed at a large distance from the objective lens of telescope, a virtual, inverted and magnified image of the object is formed at the least distance of distinct vision from the eye held close to the eye piece. Figure below shows the astronomical telescope.
Working: Parallel rays of light coming from the distant object forms real, inverted and diminished image of an object.
This image acts as an object for the eye piece. The eye piece than forms virtual, erect and magnified image of the distant object.
Question 43.An object of height 5 cm is placed of 10cm from convex mirror of focal length 15cm.Find the position nature and size of an image.
Answer:According to the question;
Object distance (u) = - 10 cm;
Focal length (f) = 15 cm Image distance = v cm;
Height of object = 5 cm
By Mirror Formula;
⇒ 
.
⇒ 
⇒ 
⇒ V = 6 cm.
Since image distance v is positive
⇒ Image is virtual and erect.
Magnification (m) = 
⇒ 
⇒
∵ v < u ⇒ Magnification < 1 hence image is diminished.
A virtual, erect and diminished image of height 3 cm is formed 6 cm behind the mirror.
Question 44.An object of height 6 cm is placed at a distance of 15cm from concave mirror of focal length 10cm.Find the position nature and height of an image.
Answer:According to the question;
Object distance (u) = - 15 cm;
Focal length (f) = - 10 cm Image distance = v cm;
Height of object = 6 cm;
By Mirror Formula;
⇒ 
.
⇒ 
⇒ 
⇒ V = - 30 cm.
Since image distance v is negative
∴ Image is real and inverted.
Magnification (m) =
⇒ 
⇒
∴ Height of image = 12 cm.
∵ v > u ⇒ Magnification > 1 hence image is magnified.
A real, inverted and magnified image of height 12 cm is formed 30 cm in front of the mirror beyond centre of curvature.
Question 45.The rays of light are entering from glass into glycerine. If the absolute refractive indices of glass and glycerine are 1.5 and 1.47 respectively, find the refracting index of glycerine relative to glass.
Answer:Refractive index (R.I) of glycerine with respect to glass =
⇒ Refractive index (R.I) of glycerine with respect to glass = 
Hence Refractive index (R.I) of glycerine with respect to glass is 0.98
Question 46.The refractive index of light entering glass from water is 1.12. Find absolute refractive index of water if the absolute refractive index of glass is 1.5.
Answer:Refractive index (R.I) of glass with respect to water = 
⇒ 1.12 = 
⇒ Absolute R.I of water = 
Hence Absolute Refractive index of water is 1.34
Question 47.When the light entering from glass to water, refractive index of water with respect to glass is 0.9.The angle of incidence at the surface separating two media is 26o48’.Find the angle of refraction at the surface.
Take sin 26o48’ = 0.45 approximately.
Answer:Refractive index (R.I) of water with respect to glass =
Given;
Angle of incidence (i) = 26o48’
Refractive index of water with respect to glass = 0.9
⇒ 0.9 = 
⇒ 
= 
⇒ 
⇒ r = 
⇒ r = 300.
Hence angle of refraction is 300.
Question 48.An object on place perpendicular to the principle axis of a convex lens having focal length 10cm.The distance of an object from the lens is 15 cm. Find the position of an image.
Answer:According to the question;
Focal length (f) = + 10 cm
Object distance (u) = - 15 cm.
Image distance = v.
By lens formula;
⇒ 
⇒ 
⇒ 
⇒ V = 30 cm.
Hence image is formed at distance of 30 cm from the lens.
Question 49.An object is placed perpendicular to the principle axis of concave lens of focal length 30cm.Find the position of an image when the object is at a distance 20cm from the lens.
Answer:According to the question;
Focal length (f) = - 30 cm
Object distance (u) = - 20 cm.
Image distance = v.
By lens formula;
⇒ 
⇒ 
⇒ 
⇒ V = - 12 cm.
Hence image is formed at distance of 12 cm from the lens.
Question 50.A power of convex lens is + 4.0 D. At what distance should the object from the lens be placed to obtain its real and inverted image of the same size on the screen?
Answer:Given;
Power of convex lens = + 4 D.
We know that;
Power (P) = 
∴ Focal length (f) = 
Hence f = 0.25 m.
Also
Given that;
Size of the object = size of the image.
For convex lens this is possible only when the object is placed at the centre of curvature.
∴ Object distance = 2f = 2 × 0.25 = 0.5 m.
0.5m = 0.5 × 100 = 50 cm.
Hence object is placed at 50 cm at centre of curvature.
What is the wavelength range of visible light?
A. 4 x 10 - 7m to 8x10 - 7 m
B. 4x10 - 9 m to 8x10 - 9 m
C. 4x10 - 5 m to 8x10 - 5 m
D. 4x10 - 6 m to 8x10 - 6 m
Answer:
Option (A): it is a fact that Wavelength of visible light is between 4 × 10 - 7 and 8 × 10 - 7. Therefore, Option A is correct.
Question 2.
What is the relation between radius of curvature (R) and the focal length (f) of a spherical mirror?
A. R = f/2
B. R = f
C. R = 2f
D. R = 3f
Answer:
Radius of curvature (R) is equal to twice the focal length that means R = 2f. Therefore, Option C is correct.
Question 3.
Which type of reflection will be represented by a light reflected from a book?
A. Regular
B. Irregular
C. Both types
D. None
Answer:
Surface of Book is rough, therefore the reflection from book is irregular because the reflected light beam will not go in the particular direction.
Since book has rough surface therefore Option B is correct.
Question 4.
Through which of the following points, a ray passing through a centre of curvature and reflected by concave mirror will pass through?
A. Focus
B. Centre of curvature
C. Pole
D. All
Answer:
When ray of light passes through the centre of curvature of the concave mirror it strikes along the normal that means it is incident on mirror at 900.Hence the incident ray coincides with the normal and retraces its path i.e. passes through centre of curvature.
As other options don’t go by above fact, so they are incorrect.
Question 5.
At what distance in front of the concave mirror should an object be placed to get its virtual and erect image?
A. At centre of curvature
B. Beyond centre of curvature
C. Between focus and pole
D. At focus
Answer:
when an object is placed between the focus and pole of concave mirror the image is formed behind the concave mirror and is virtual, erect and enlarged. As shown in figure below.
Option (A): When object is kept at centre of curvature than image is formed at centre of curvature and is real, inverted. Therefore, Option A is incorrect.
Option (B): when an object is kept beyond centre of curvature the image formed is real, inverted and diminished. Therefore, Option B is incorrect.
Option (C): when an object is placed between focus and pole than image is virtual and erect as explained in explanation. Therefore, Option C is correct.
Option (D): when an object is placed at focus than image is formed at infinity and is real. Therefore, Option D is incorrect.
Question 6.
The magnification of plane mirror is always
A. More than 1
B. 1
C. Less than 1
D. Zero
Answer:
Magnification =
And plane mirror forms the image of same size as that of object. (Height of Image = Height of Object).
Therefore, Magnification is 1
Question 7.
The focal length of plane mirror is......
A. Zero
B. Infinity
C. Uncertain
D. Equal to object distance.
Answer:
A plane mirror can be considered as a spherical mirror of infinite radius of curvature. And Focal length is half of radius of curvature therefore it is infinity.
Question 8.
The distance between the object at 2 m from a plane mirror and its image is....
A. 4 m
B. 1 m
C. 2m
D. 3m
Answer:
In plane mirror Object distance from the mirror is equal to the image distance from the mirror.
Since Object distance from mirror = 2 m
⇒ Image distance from mirror = 2 m.
∴ Distance between image and object = 2 m + 2 m = 4 m.
Question 9.
At what distance should an object be placed to obtain its real, inverted and of same height as the objet by a convex lens?
A. At focus
B. Between focus and centre of curvature
C. At centre of curvature
D. Between optical centre and focus.
Answer:
when an object is kept at centre of curvature of convex lens same size, real and inverted image of an object is formed at centre of curvature as shown in figure below.
Option (A): when an object is kept at focus than image is formed at infinity and is magnified. Therefore, Option A is incorrect.
Option (B): when an object is kept between focus and centre of curvature than image is formed on other side of lens and is magnified (enlarged). Therefore, Option B is incorrect.
Option (C): when an object is kept at centre of curvature than real, inverted image is formed at centre of curvature as explained in explanation. Therefore, option C is correct.
Option (D): when an object is kept between optical centre and focus the image formed is virtual and erect. Therefore, Option D is incorrect.
Question 10.
Which of the following materials has maximum optical density?
A. Glass
B. Water
C. Pearl
D. Diamond
Answer:
optical density is defined as the degree to which the refractive medium refracts the transmitted ray of light. Therefore, Medium having highest refractive index will have highest optical density.
For refractive indices,
Water (1.33) < pearl (1.46) < glass (1.52) < diamond (2.42)
Question 11.
The absolute refractive index of any medium is always...
A. 1
B. >1
C. <1
D. Zero
Answer:
Absolute refractive index of medium is defined as , and speed of light in vacuum is highest than its speed in any medium. Therefore, n>1.
Question 12.
Which of the lenses with focal length 10cm, 20cm, 25cm and 50cm has maximum power?
A. 50 cm
B. 25 cm
C. 20 cm
D. 10cm
Answer:
Power of lens = .Therefore if focal length is less power is more and if focal length is more power is less.
Option (A): 50 cm is the highest focal length in given options; hence it will have lowest power. Therefore, Option A is incorrect.
Option (B): 25 cm is not the lowest focal length; hence it will not have highest power. Therefore, Option B is incorrect.
Option (C): 20 cm is not the lowest focal length; hence it will not have highest power. Therefore, Option C is incorrect.
Option (D): 10 cm is the lowest focal length in given options. Hence it will have highest power. Therefore, Option D is correct.
Question 13.
What is the focal length of a convex lens having power + 5.0D?
A. - 10cm
B. - 20cm
C. + 10cm
D. + 20cm
Answer:
Power of lens = .
⇒ Focal length =
1m = 100 cm.
⇒ 0.2 m = 0.2 × 100 = 20 cm.
∴ Focal length = 20 cm. (Positive for convex lens.)
Option (A): since focal length is positive for convex lens. Therefore, Option A is incorrect.
Option (B): since focal length is positive for convex lens. Therefore, Option B is incorrect.
Option (C): focal length is 20 cm not 10 cm therefore option (C) is incorrect.
Option (D): as explained in explanation focal length = + 20 cm therefore option D is correct.
Question 14.
If the absolute refractive indices of water, benzene, and sapphire are 1.33, 1.50 and 1.77 respectively, then which medium has maximum relative refractive index?
A. Sapphire relative to water
B. Sapphire relative to benzene
C. Benzene relative to water
D. Water relative to benzenes
Answer:
The relative refractive index between a pair of media is the ratio of their absolute refractive indices.
Let n1 = absolute refractive index of water; n2 = absolute refractive index of benzene; n3 = absolute refractive index of sapphire.
Thus,
It can be seen that n31 is maximum and corresponds to refractive index of sapphire relative to water.
Question 15.
Which type of an image is formed by a plane mirror?
A. Real and inverted
B. Real and erect
C. Virtual and erect
D. Virtual and inverted
Answer:
Option (A): Plane mirror never forms real and inverted image, therefore option (A) is incorrect.
Option (B): the image formed by plane mirror is virtual and erect it is not real; therefore option (B) is incorrect.
Option (C): the plane mirror always form the virtual, erect image of same size as that of an object, therefore (C) is the correct answer.
Option (D): image formed is erect not inverted; therefore (D) is not the correct answer.
Question 16.
If the absolute refractive indices of water and glass are 4/3 and 3/2 respectively, then what will be the ratio of velocity of light in water to that of glass?
A. 2
B. 8/9
C. 9/8
D. 1/2
Answer:
We know that
∴
×
⇒
⇒
Question 17.
The absolute refractive indices of water glass and diamond are 1.77, 1.50 and 2.42 respectively, which medium is most optically denser?
A. Water
B. Glass
C. Diamond
D. None
Answer:
optical density is defined as the degree to which the refractive medium refracts the transmitted ray of light. Therefore, Medium having highest refractive index will have highest optical density.
Question 18.
Which of the following always form virtual image?
A. Concave mirror and convex lens
B. Convex mirror and concave lens
C. Convex mirror and convex lens
D. Concave mirror and Concave lens
Answer:
Option (A): Concave mirror and convex lens
Forms both virtual as well as real images, hence A is not the correct answer.
Option (B): for all positions of object convex mirror and concave lens always forms virtual and erect images behind the mirror/lens, therefore B is the correct answer.
Option (C): convex lens not forms only virtual images it also forms real images, hence C is not the correct answer.
Option (D): Concave mirror forms both real and virtual images, hence D is not the correct answer.
Question 19.
What will be the angle of refraction for the light ray incident normal at the surface?
A. 900
B. 600
C. 300
D. 00
Answer:
since the light ray is incident normally to the surface, therefore it will pass without any refraction. Hence angle of refraction = 00.
Question 20.
The compound microscope consists of two convex lenses of 5 cm and 20 cm focal length, and then which of them will be object lens and eye piece?
A. Object lens with 20cm focal length and eye piece with 5 cm focal length.
B. Object lens with 5 cm focal length and eye piece with 20 cm focal length.
C. Both should have 20 cm focal length.
D. Both should have 5 cm focal length.
Answer:
For the compound microscope focal length of an objective lens is smaller than eye piece lens.f0 = objective lens focal length, fe = eye piece focal length.
⇒ f0 < fe
Question 21.
What is called regular and irregular reflection of light?
Answer:
Regular reflection: when a parallel beam of light is incident on a plane or smooth surface, a beam remains parallel after reflection in a specific direction. Such reflection is called regular reflection. The figure below illustrates regular reflection.
Example: Reflection of light by plane mirror.
Irregular reflection: when a parallel beam of light is incident on rough or irregular surface like Book, stone etc., the beam does not remains parallel but spreads over wide region after reflection. Such reflection is called irregular reflection. The figure below illustrates irregular reflection.
Example: we can see book, table, and chair etc. due to irregular Reflection of light these objects.
Question 22.
Write the laws of reflection of light.
Answer:
the laws of reflection of light is as follows.
The angle of incidence is equal to the angle of reflection that means .The figure below illustrates it.
The incident ray, reflected ray, and normal to the mirror at the point of incidence all lie in the same plane.
The figure below illustrates the law more clearly.
Question 23.
What are called centre of curvature and radius of curvature of mirror?
Answer:
Centre of curvature: the centre of a spherical shell from which the mirror is made is called as the Centre of curvature of mirror. It is denoted by ‘C’.
Radius of curvature: the radius of a spherical shell from which the mirror is made is called as the radius of curvature of mirror. It is denoted by ‘R’.
The figure below shows the centre and radius of curvature of concave lens.
Question 24.
Draw a ray diagram showing position, nature and size of an image formed by concave mirror when the object is placed beyond the centre of curvature.
Answer:
Ray Diagram:
Position of Image: Between Centre of curvature (C) and focus (F).
Nature of image: real and inverted.
Size of image: diminished.
Question 25.
Draw a ray - diagram showing position, nature and size of an image formed by concave mirror when the object is placed between pole and principle focus.
Answer:
Ray Diagram:
Position of Image: behind the mirror
Nature of image: virtual and erect
Size of image: Magnified (enlarged).
Question 26.
Draw a ray - diagram showing position, nature and size of an image formed by convex mirror when the object is placed between infinite distance and pole.
Answer:
Ray Diagram:
Position of Image: between pole (P) and principal focus (F) behind the mirror.
Nature of image: virtual and erect.
Size of image: diminished.
Question 27.
Obtain the position, nature and size of an image formed by a plane mirror from the formula of magnification.
Answer:
In case of plane mirror the magnification is + 1;
Positive value of magnification indicates that image formed by plane mirror is erect.
Also, Magnification =
⇒ Image height = Object height.
Therefore, Image formed by plane mirror is of the same size as of the object.
Magnification =
⇒ v = - u;
This shows that image formed by plane mirror is at same distance as the object but behind the mirror means image is virtual.
Thus, Image formed by plane mirror is virtual, erect and same as size of the object.
Question 28.
Write laws of refraction of light.
Answer:
The refraction of light occurs according to the following laws which are as follows.
The incident ray, refracted ray, and the normal to the surface separating two media at the point of incidence all lie in the same plane.
The ratio of sine of angle of incidence and sine of angle of refraction remains constant ( = constant) subject to certain conditions. This law is called Snell’s law of refraction.
Question 29.
What is called the absolute refractive index of a medium? Obtain the general from of Snell’s law in terms of refractive indices of two media?
Answer:
The refractive index of medium with respect to the vacuum is called absolute refractive index.
Snell’s law in terms of refractive indices of two media: Let,
N1 = absolute refractive index of medium 1.
N2 = absolute refractive index of medium 2.
V1 = velocity of light in medium 1.
V2 = velocity of light in medium 2.
C = velocity of light in vacuum.
Then
N1 = and N2 =
∴
⇒ N21 = ………….. (1)
Now According to Snell’s law
N21 = …………….. (2)
From (1) and (2)
This is the general form of Snell’s Law.
Question 30.
Draw a ray - diagram showing the position, nature and size of an image formed by convex lens when the object is placed at centre of curvature of lens.
Answer:
Ray Diagram:
Position of Image: on other side of lens at centre of curvature.
Nature of image: real and inverted
Size of image: Same as size of object.
Question 31.
Draw a ray - diagram showing the position, nature and size of an image formed by convex lens when the object is placed between its optical centre and focus.
Answer:
Ray Diagram:
Position of Image: beyond the centre of curvature (2F) on the same side of lens where object is kept
Nature of image: virtual and erect
Size of image: Magnified (enlarged).
Question 32.
Draw a ray - diagram showing the position, nature and size of an image formed by concave lens when the object is placed between at optical centre and infinite point.
Answer:
Ray Diagram:
Position of Image: between focus (F) and optical centre (C) on the same side of lens where object is kept.
Nature of image: virtual and erect
Size of image: small (diminished).
Question 33.
What is called the magnification of an image? Derive the formula of magnification for spherical lens.
Answer:
Magnification of an image is defined as the ratio of the height of image to the height of object.
Magnification (m) =
In the figure above by geometry we can easily see that ΔABO and ΔA’B’O’ are similar.
∴ …………… (1)
From definition of magnification.
Magnification (m) =
⇒ ……… (2)
From (1) and (2)
Magnification = which is the formula for magnification of lens.
Question 34.
Explain the reflection by a plane mirror by drawing suitable figure.
Answer:
MM’ = plane mirror.
AB = object of size ‘h’ on the left - hand side of the mirror at a distance ‘u’.
AP is the incident ray of light from point A of the object which falls on the mirror at point P. This incident ray is reflected back in the same path PA as shown in the figure above.
Another incident ray OC falls on the mirror at a point O and is reflected along the path OC.
Now, since reflected rays PA and OC are diverging and therefore cannot meet each other in front of the mirror, hence we extend these rays behind the mirror by dotted lines as shown in the figure below.
On extending these rays behind the mirror, we see that these rays meet at point A’ at a distance V’. Therefore A’ is the virtual image of point A of the object AB.
Similarly, a virtual image of point B will be formed behind the mirror as B’.
Now, to get a complete image of the object AB, we join the point A and B to point A’ and B’ by a dotted line.
We find that the image A’B’ being formed is virtual, erect and of same shape and size as the object AB; thereby giving us the characteristics of images formed by the plane mirror.
Question 35.
Give the Cartesian sign convention for the reflection by spherical mirror.
Answer:
Pole (P) of the mirror is taken as the origin and the Principal axis is taken as the horizontal X - axis. The sign conventions are as follows:
The object is always placed on the left side of the mirror.
All the distances measured to the right side of the origin are taken as positive, while the distances measured to the left side of the origin are taken as negative.
All the distances parallel to the principal axis of the mirror are measured from the pole of the mirror.
Distances measured vertically upwards from the principal axis of the mirror are taken as positive.
Distances measured vertically downwards from the principal axis of the mirror are taken as negative.
The focal length of Concave mirror is negative while that of convex mirror is positive.
The figure below illustrates the conventions used.
Question 36.
With the necessary figure, explain the refraction of light through a rectangular glass slab.
Answer:
Consider the rectangular slab as shown in the figure below.
AE = incident ray;
i = Angle if incidence;
r = Angle of refraction;
On entering the glass slab the light ray bends towards the normal and travels along EF.
The refracted ray EF is incident on the face SR at angle of incidence r’.
The emergent ray FD bends away from normal at angle of refraction e.
The emergent ray FD is parallel to the incident ray AE but it has been laterally displaced with respect to incident ray.
Question 37.
Obtain the lens formula for spherical lens.
Answer:
Consider the image formation by convex lens as shown in the figure below.
By Cartesian sign convention;
Object distance (OB) = - u;
Image distance (OB’) = + v;
Focal length (OF2 = OF2) = f
From the geometry we can see that in the figure above right triangle ΔABO and ΔA’B’O’ are similar.
∴ …… (1)
Also, in figure above by geometry right triangle ΔODF2 and ΔBAF2 are similar.
∴
∴ (∵ OD = AB, opposite sides of rectangle ABOD are equal).
∴
∴ ………. (2)
From (1) and (2)
⇒
⇒ .
Dividing each term by uvf, we get.
⇒
⇒
This equation is called the ‘Lens Formula’
Question 38.
Explain how the position of images is located for spherical mirror by considering the different rays using necessary ray - diagrams.
Answer:
The image formed by spherical mirrors can be located by constructing ray diagrams. The point of intersection of two rays will give the position of the image.
Following rays can be taken into consideration while locating the image.
1. A ray parallel to the principal axis after reflection passes through the focus or appears to diverge from the focus. As shown in the figure below.
2. A ray passing through the principal focus of the concave mirror or ray which is directed towards the principal focus of convex mirror will emerge parallel to the principal axis as shown in the figure below.
3. A ray passing through the centre of curvature of the concave mirror or directed towards the centre of curvature of convex mirror after reflection is, reflected along the same path, as shown in the figure below.
4. A light ray incident obliquely to the principal axis towards the pole of a concave mirror or convex mirror is reflected obliquely following the laws of reflection. The figure below shows it.
Question 39.
Write a note on power of lens.
Answer:
Power of lens is defined as the efficiency with which a lens can converge or diverge the light ray.
The reciprocal of focal length of the lens is called the power of the lens (P).
P (in dioptre) =
The S.I unit of power of lens is dioptre denoted by D.
Power of convex lens is taken as positive and of concave lens is negative. The instrument used for measuring the power of lens is called dioptremeter.
Question 40.
Derive the formula for spherical mirror
Answer:
According to the Cartesian system in the above figure.
Object distance (PB) = - u
Image distance (PB’) = - v
Focal length (PF) = - f
Radius of curvature (PC) = - R
From the geometry we can see that in the figure above right triangle ΔABP and ΔA’B’P’ are similar.
∴
∴ ……. (1)
Also ΔABC and ΔA’B’C’ are similar.
∴ ……. (2)
Also
From above figure
CB’ = PC – PB’ = - R - (- v) = - R + v.
And
CB = PB – PC = - u - (- R) = - - - u + R.
From (2)
∴
⇒ R (u + v) = 2uv
Dividing both sides by Ruv we get.
We know that f =
Putting this in the above equation we get
This equation is the ‘Mirror Formula’
Question 41.
Explain the construction and working of a compound microscope with a neat ray - diagram.
Answer:
Construction: A compound microscope consists of two convex lenses. The lens towards object is called objective lens and the lens near eye is called the eye piece. The focal length of objective lens is less than the focal length of eye piece. This combination forms the magnified image of the object.
The figure below shows the compound microscope.
Working: The object AB is placed in front of objective lens in such a way that its real, inverted and magnified image A’B’ is obtained beyond the centre of curvature of objective lens.
The image A’B’ becomes an object for the eye piece. The position of image is adjusted such that it lies within the focal length of eye piece.
The eye piece forms the virtual, erect and magnified image A”B” of the object.
Question 42.
Write a note on astronomical telescope.
Answer:
An astronomical telescope is an optical instrument which is used to see the magnified image of distant heavenly bodies like stars, planets, satellites and galaxies etc.
The final image formed by an astronomical telescope is always virtual, inverted and magnified.
Principle: An astronomical telescope works on the principle that when an object to be magnified is placed at a large distance from the objective lens of telescope, a virtual, inverted and magnified image of the object is formed at the least distance of distinct vision from the eye held close to the eye piece. Figure below shows the astronomical telescope.
Working: Parallel rays of light coming from the distant object forms real, inverted and diminished image of an object.
This image acts as an object for the eye piece. The eye piece than forms virtual, erect and magnified image of the distant object.
Question 43.
An object of height 5 cm is placed of 10cm from convex mirror of focal length 15cm.Find the position nature and size of an image.
Answer:
According to the question;
Object distance (u) = - 10 cm;
Focal length (f) = 15 cm Image distance = v cm;
Height of object = 5 cm
By Mirror Formula;
⇒ .
⇒
⇒
⇒ V = 6 cm.
Since image distance v is positive
⇒ Image is virtual and erect.
Magnification (m) =
⇒
⇒
∵ v < u ⇒ Magnification < 1 hence image is diminished.
A virtual, erect and diminished image of height 3 cm is formed 6 cm behind the mirror.
Question 44.
An object of height 6 cm is placed at a distance of 15cm from concave mirror of focal length 10cm.Find the position nature and height of an image.
Answer:
According to the question;
Object distance (u) = - 15 cm;
Focal length (f) = - 10 cm Image distance = v cm;
Height of object = 6 cm;
By Mirror Formula;
⇒ .
⇒
⇒
⇒ V = - 30 cm.
Since image distance v is negative
∴ Image is real and inverted.
Magnification (m) =
⇒
⇒
∴ Height of image = 12 cm.
∵ v > u ⇒ Magnification > 1 hence image is magnified.
A real, inverted and magnified image of height 12 cm is formed 30 cm in front of the mirror beyond centre of curvature.
Question 45.
The rays of light are entering from glass into glycerine. If the absolute refractive indices of glass and glycerine are 1.5 and 1.47 respectively, find the refracting index of glycerine relative to glass.
Answer:
Refractive index (R.I) of glycerine with respect to glass =
⇒ Refractive index (R.I) of glycerine with respect to glass =
Hence Refractive index (R.I) of glycerine with respect to glass is 0.98
Question 46.
The refractive index of light entering glass from water is 1.12. Find absolute refractive index of water if the absolute refractive index of glass is 1.5.
Answer:
Refractive index (R.I) of glass with respect to water =
⇒ 1.12 =
⇒ Absolute R.I of water =
Hence Absolute Refractive index of water is 1.34
Question 47.
When the light entering from glass to water, refractive index of water with respect to glass is 0.9.The angle of incidence at the surface separating two media is 26o48’.Find the angle of refraction at the surface.
Take sin 26o48’ = 0.45 approximately.
Answer:
Refractive index (R.I) of water with respect to glass =
Given;
Angle of incidence (i) = 26o48’
Refractive index of water with respect to glass = 0.9
⇒ 0.9 =
⇒ =
⇒
⇒ r =
⇒ r = 300.
Hence angle of refraction is 300.
Question 48.
An object on place perpendicular to the principle axis of a convex lens having focal length 10cm.The distance of an object from the lens is 15 cm. Find the position of an image.
Answer:
According to the question;
Focal length (f) = + 10 cm
Object distance (u) = - 15 cm.
Image distance = v.
By lens formula;
⇒
⇒
⇒
⇒ V = 30 cm.
Hence image is formed at distance of 30 cm from the lens.
Question 49.
An object is placed perpendicular to the principle axis of concave lens of focal length 30cm.Find the position of an image when the object is at a distance 20cm from the lens.
Answer:
According to the question;
Focal length (f) = - 30 cm
Object distance (u) = - 20 cm.
Image distance = v.
By lens formula;
⇒
⇒
⇒
⇒ V = - 12 cm.
Hence image is formed at distance of 12 cm from the lens.
Question 50.
A power of convex lens is + 4.0 D. At what distance should the object from the lens be placed to obtain its real and inverted image of the same size on the screen?
Answer:
Given;
Power of convex lens = + 4 D.
We know that;
Power (P) =
∴ Focal length (f) =
Hence f = 0.25 m.
Also
Given that;
Size of the object = size of the image.
For convex lens this is possible only when the object is placed at the centre of curvature.
∴ Object distance = 2f = 2 × 0.25 = 0.5 m.
0.5m = 0.5 × 100 = 50 cm.
Hence object is placed at 50 cm at centre of curvature.