GEOMETRICAL OPTICS
OBJECTIVE-I
1. A point source of light is placed in front of a plane mirror.
(a) All the reflected rays meet at a point when produced backward.
(b) Only the reflected rays close to the normal meet at a point when produced backward.
(c) Only the reflected rays making a small angle with the mirror, meet at a point when produced backward.
(d) Light of different colors makes different images.
ANSWER: (a)
EXPLANATION: See the diagram below,
Diagram for Q-1
Take any ray PR which is reflected at Q and QR is the reflected ray. Since the angle of incidence is equal to the angle of reflection,
∠PQN =∠RQN=∠OPQ =α
RQ, when produced back, meets PO produced at P'.
In right-angled triangles PQO and P'QO,
∠QPO =∠QP'O =α
Thus both are equal in all respect.
So, PO =P'O.
For any other ray, it can be proved.
Hence the option (a).
P' becomes a virtual image of P.
2. Total internal reflection can take place only if
(a) Light goes from optically rarer medium (smaller refractive index) to optically denser medium.
(b) light goes from optically denser medium to rarer medium
(c) the refractive indices of the two media are close to each other
(d) the refractive indices of the two media are widely different.
ANSWER: (b)
EXPLANATION: When the light goes from a denser medium to lighter medium, it goes away from the normal. When the angle of incidence is increased, the angle of refraction becomes 90°. It is the critical angle. Now if the angle is increased further, the phenomenon of refraction is changed to total internal reflection.
3. In image formation from spherical mirrors, only paraxial rays are considered because they are
(a) easy to handle geometrically
(b) contain most of the intensity of the incident light
(c) form nearly a point image of a point source
(d) show a minimum dispersion effect.
ANSWER: (c)
EXPLANATION: The reflected rays from a point source in a spherical mirror do not meet at a single point. To avoid it, rays close to the principal axis (Paraxial rays) are taken. It forms nearly a point image of a point source.
4. A point object is placed at a distance of 30 cm from a convex mirror of focal length 30 cm. The image will form at
(a) infinity
(b) pole
(c) focus
(d) 15 cm behind the mirror.
ANSWER: (d)
EXPLANATION: The option can be selected by suitable reasoning. In a convex mirror, the object at infinity has its image at the focus behind the mirror. So option (c) is not correct because the object here is not at infinity.
The image in a convex mirror never forms at infinity. The option (a) is not correct.
The image will be near the pole if the object is near the pole, which is not the case here. The option (b) is not correct.
The only option left is (d) and the image position is between focal length and the pole for the object between infinity and the pole. Hence the option (d) is correct.
5. Figure (19-Q2) shows two rays A and B being reflected by a mirror and going as A' and B'. The mirror
(a) is plane
(b) is convex
(c) is concave
(d) may be any spherical mirror.
Figure for Q-5
ANSWER: (a)
EXPLANATION: The parallel rays after reflection are still parallel, which is possible only in a plane mirror. Concave or convex mirrors either converge or diverge the parallel rays. Hence the option (a) is correct.
6. The image formed by a concave mirror
(a) is always real
(b) is always virtual
(c) is certainly real if the object is virtual
(d) is certainly virtual if the object is real.
ANSWER: (c)
EXPLANATION: The image formed by a concave mirror may be real or virtual depending upon the position of the object. Hence the option (a) and (b) are wrong. Only the real object placed between the focus and the pole has a virtual image otherwise it is real. The option (d) is not correct.
Hence the only option remaining is (c).
7. Figure (18-Q3) shows three transparent media of refractive indices µ₁, µ₂, and µ₃. A point object O is placed in the medium µ₂. If the entire medium on the right of the spherical surface has a refractive index µ₁, the image forms at O'. If this entire medium has refractive index µ₃, the image forms at O". In the situation shown,
(a) the image forms between O' and O".
(b) the image forms to the left of O'.
(c) the image forms to the right of O".
(d) two images form, one at O' and the other at O".
Figure for Q-7
ANSWER: (d)
EXPLANATION: The rays incident on the surface above the principal axis will be refracted in the medium of R.I. µ₃ and form the image at O". But the rays falling below the axis will be refracted in the medium having R.I. µ₁ and the image will be formed at O'. Hence two images will be formed.
8. Four modifications are suggested in the lens formula to include the effect of the thickness t of the lens. Which one is likely to be correct?
(a) 1/v - 1/u = t/uf
(b) t/v² - 1/u = 1/f
(c) 1/(v-t) - 1/(u+t) = 1/f
(d) 1/v - 1/u +t/uv = t/f
ANSWER: (c)
EXPLANATION: The thickness of the lens will affect the object distance or image distance. Either the thickness t will be added or subtracted to u or v. It will not be in a proportion as shown in the options (a), (b) and (d). Hence the option (c) is correct.
9. A double convex lens has two surfaces of equal radii R and refractive index µ = 1.5, we have,
(a) f =R/2
(b) f =R
(c) f = -R
(d) f =2R
ANSWER: (b)
EXPLANATION: From lens makers formula
1/f = (µ-1){1/R - 1/R'}
here µ = 1.5, R' = -R
→1/f = 0.5*2/R =1/R
→f = R
Hence the option (b) is true.
10. A point source of light is placed at a distance of 2f from a converging lens of focal length f. The intensity on the other side of the lens is maximum at a distance
(a) f
(b) between f and 2f
(c) 2f
(d) more than 2f.
ANSWER: (c)
EXPLANATION: The intensity of the light on the other side is maximum where the image of the point source is formed.
Here u = -2f, f = f, hence from the lens formula
1/v -1/(-2f) = 1/f
→1/v = 1/f -1/2f = 1/2f
→v = 2f
Hence the image will be formed at 2f distance from the lens on the other side.
11. A parallel beam of light is incident on a converging lens parallel to its principal axis. As one moves away from the lens on the other side on its principal axis, the intensity of light
(a) remains constant
(b) continuously increases
(c) continuously decreases
(d) first increases then decrease.
ANSWER: (d)
EXPLANATION: The parallel beam of light converges at the focus on the other side and then diverges. Hence till the person moves to the focus the intensity will increase. Beyond the focus, it will decrease. Hence the option (d).
12. A symmetric double convex lens is cut in two equal part by a plane perpendicular to the principal axis. If the power of the original lens was 4D, the power of a cut lens will be
(a) 2D
(b) 3D
(c) 4D
(d) 5D
ANSWER: (a)
EXPLANATION: Let the radius of curvature of the symmetric double convex lens be R each. A plane perpendicular to the principal axis that cuts the lens in equal parts, makes each part a Plano-convex lens as shown in the figure,
Diagram for Q-12
If the focal length of the original lens = f
1/f = (µ-1){1/R-1/(-R)} =2(µ-1)/R
Hence the power of the original lens
P = 4D =1/f = 2(µ-1)/R
→(µ-1)/R = 4D/2 =2D
For the part lens R=R and R' = - ∞, if f' = focal length, then
1/f' = (µ-1){1/R -1/(-∞)} =(µ-1)/R
Hence the power of the part lens P' =1/f'
→P' = (µ-1)/R =2D
Hence the option (a).
13. A symmetric double convex lens is cut in two equal part by a plane containing the principal axis. If the power of the original lens was 4D, the power of a cut lens will be
(a) 2D
(b) 3D
(c) 4D
(d) 5D
ANSWER: (c)
EXPLANATION: The plane containing the principal axis will cut the lens in two semicircular pieces and each piece will be a lens having the same radius of the curvatures. Hence each of the pieces will have the same focal length and the same power. Thus the option (c).
14. Two concave lenses L₁ and L₂ are kept in contact with each other. If the space between the two lenses is filled with a material of refractive index µ = 1, the magnitude of the focal length of the combination
(a) becomes undefined
(b) remains unchanged
(c) increases
(d) decreases
ANSWER: (c)
EXPLANATION: The focal length F of the combination is given as
1/F = 1/f₁ + 1/f₂
It is like resistors connected in parallel. The numerical value of equivalent resistance is less than the smallest resistor. Similarly, the numerical value of the equivalent focal length will decrease. Since the 1/F is the power of the lens and in the case of a diverging lens, it is negative as in this case. Hence F is also negative. If the numerical value of F decreases that means its negative will increase.
15. A thin lens is made with a material having a refractive index µ = 1.5. Both the sides are convex. It is dipped in water (µ = 1.33). It will behave like
(a) a convergent lens
(b) a divergent lens
(c) a rectangular slab
(d) a prism
ANSWER: (a)
EXPLANATION: The double convex lens behaves as a convergent lens in the air because the µ of the lens is greater than air. When it is dipped in water its µ is still greater than water hence it will act as a convergent lens.
16. A convex lens is made of a material having refractive index 1.2, both the surfaces of the lens are convex. If it is dipped into water (µ = 1.33). It will behave like
(a) a convergent lens
(b) a divergent lens
(c) a rectangular slab
(d) a prism
ANSWER: (b)
EXPLANATION: The µ of the convex lens is greater than air so it will act as a convergent lens in the air. When it is dipped in water its µ is less than water. Now the rays entering and exiting the lens will bend toward or away from the normal in an opposite fashion than when the lens was placed in air. So the convergent lens in the air will now behave as a divergent lens in the water.
17. A point object O is placed on the principal axis of a convex lens of focal length f = 20 cm at a distance of 40 cm to the left of it. The diameter of the lens is 10 cm. An eye is placed 60 cm to the right of the lens and a distance h below the principal axis. The maximum value of h to see the image is
(a) 0
(b) 2.5 cm
(c) 5 cm
(d) 10 cm.
ANSWER: (b)
EXPLANATION: Let us draw a diagram as below,
Diagram for Q-17
Since the object is placed at 2f distance, the image will also be at 2f = 40 cm distance at I. The eye is at E. AC = 5 cm, CI = 40 cm, ID = 20 cm and DE = h. Since triangles ACI and DEI are similar triangles their sides will be proportional. Hence
DE/AC = ID/CI
→h/5 = 20/40 =1/2
→h = 5/2 =2.5 cm
18. The rays of different colors fail to converge at a point after going through a converging lens. This defect is called
(a) spherical aberration
(b) distortion
(c) coma
(d) chromatic aberration
ANSWER: (d)
EXPLANATION: Since the refractive index of a medium is different for different colors. Hence the rays of different colors fail to converge at a point. Since this aberration is related to colored rays it is called a chromatic aberration because chrome is a Greek word for color.
===<<<O>>>===
1. A point source of light is placed in front of a plane mirror.
(a) All the reflected rays meet at a point when produced backward.
(b) Only the reflected rays close to the normal meet at a point when produced backward.
(c) Only the reflected rays making a small angle with the mirror, meet at a point when produced backward.
(d) Light of different colors makes different images.
ANSWER: (a)
EXPLANATION: See the diagram below,
Take any ray PR which is reflected at Q and QR is the reflected ray. Since the angle of incidence is equal to the angle of reflection,
∠PQN =∠RQN=∠OPQ =α
RQ, when produced back, meets PO produced at P'.
In right-angled triangles PQO and P'QO,
∠QPO =∠QP'O =α
Thus both are equal in all respect.
So, PO =P'O.
For any other ray, it can be proved.
Hence the option (a).
P' becomes a virtual image of P.
2. Total internal reflection can take place only if
(a) Light goes from optically rarer medium (smaller refractive index) to optically denser medium.
(b) light goes from optically denser medium to rarer medium
(c) the refractive indices of the two media are close to each other
(d) the refractive indices of the two media are widely different.
ANSWER: (b)
EXPLANATION: When the light goes from a denser medium to lighter medium, it goes away from the normal. When the angle of incidence is increased, the angle of refraction becomes 90°. It is the critical angle. Now if the angle is increased further, the phenomenon of refraction is changed to total internal reflection.
3. In image formation from spherical mirrors, only paraxial rays are considered because they are
(a) easy to handle geometrically
(b) contain most of the intensity of the incident light
(c) form nearly a point image of a point source
(d) show a minimum dispersion effect.
ANSWER: (c)
EXPLANATION: The reflected rays from a point source in a spherical mirror do not meet at a single point. To avoid it, rays close to the principal axis (Paraxial rays) are taken. It forms nearly a point image of a point source.
4. A point object is placed at a distance of 30 cm from a convex mirror of focal length 30 cm. The image will form at
(a) infinity
(b) pole
(c) focus
(d) 15 cm behind the mirror.
ANSWER: (d)
EXPLANATION: The option can be selected by suitable reasoning. In a convex mirror, the object at infinity has its image at the focus behind the mirror. So option (c) is not correct because the object here is not at infinity.
The image in a convex mirror never forms at infinity. The option (a) is not correct.
The image will be near the pole if the object is near the pole, which is not the case here. The option (b) is not correct.
The only option left is (d) and the image position is between focal length and the pole for the object between infinity and the pole. Hence the option (d) is correct.
5. Figure (19-Q2) shows two rays A and B being reflected by a mirror and going as A' and B'. The mirror
(a) is plane
(b) is convex
(c) is concave
(d) may be any spherical mirror.
ANSWER: (a)
EXPLANATION: The parallel rays after reflection are still parallel, which is possible only in a plane mirror. Concave or convex mirrors either converge or diverge the parallel rays. Hence the option (a) is correct.
6. The image formed by a concave mirror
(a) is always real
(b) is always virtual
(c) is certainly real if the object is virtual
(d) is certainly virtual if the object is real.
ANSWER: (c)
EXPLANATION: The image formed by a concave mirror may be real or virtual depending upon the position of the object. Hence the option (a) and (b) are wrong. Only the real object placed between the focus and the pole has a virtual image otherwise it is real. The option (d) is not correct.
Hence the only option remaining is (c).
7. Figure (18-Q3) shows three transparent media of refractive indices µ₁, µ₂, and µ₃. A point object O is placed in the medium µ₂. If the entire medium on the right of the spherical surface has a refractive index µ₁, the image forms at O'. If this entire medium has refractive index µ₃, the image forms at O". In the situation shown,
(a) the image forms between O' and O".
(b) the image forms to the left of O'.
(c) the image forms to the right of O".
(d) two images form, one at O' and the other at O".
ANSWER: (d)
EXPLANATION: The rays incident on the surface above the principal axis will be refracted in the medium of R.I. µ₃ and form the image at O". But the rays falling below the axis will be refracted in the medium having R.I. µ₁ and the image will be formed at O'. Hence two images will be formed.
8. Four modifications are suggested in the lens formula to include the effect of the thickness t of the lens. Which one is likely to be correct?
(a) 1/v - 1/u = t/uf
(b) t/v² - 1/u = 1/f
(c) 1/(v-t) - 1/(u+t) = 1/f
(d) 1/v - 1/u +t/uv = t/f
ANSWER: (c)
EXPLANATION: The thickness of the lens will affect the object distance or image distance. Either the thickness t will be added or subtracted to u or v. It will not be in a proportion as shown in the options (a), (b) and (d). Hence the option (c) is correct.
9. A double convex lens has two surfaces of equal radii R and refractive index µ = 1.5, we have,
(a) f =R/2
(b) f =R
(c) f = -R
(d) f =2R
ANSWER: (b)
EXPLANATION: From lens makers formula
1/f = (µ-1){1/R - 1/R'}
here µ = 1.5, R' = -R
→1/f = 0.5*2/R =1/R
→f = R
Hence the option (b) is true.
10. A point source of light is placed at a distance of 2f from a converging lens of focal length f. The intensity on the other side of the lens is maximum at a distance
(a) f
(b) between f and 2f
(c) 2f
(d) more than 2f.
ANSWER: (c)
EXPLANATION: The intensity of the light on the other side is maximum where the image of the point source is formed.
Here u = -2f, f = f, hence from the lens formula
1/v -1/(-2f) = 1/f
→1/v = 1/f -1/2f = 1/2f
→v = 2f
Hence the image will be formed at 2f distance from the lens on the other side.
11. A parallel beam of light is incident on a converging lens parallel to its principal axis. As one moves away from the lens on the other side on its principal axis, the intensity of light
(a) remains constant
(b) continuously increases
(c) continuously decreases
(d) first increases then decrease.
ANSWER: (d)
EXPLANATION: The parallel beam of light converges at the focus on the other side and then diverges. Hence till the person moves to the focus the intensity will increase. Beyond the focus, it will decrease. Hence the option (d).
12. A symmetric double convex lens is cut in two equal part by a plane perpendicular to the principal axis. If the power of the original lens was 4D, the power of a cut lens will be
(a) 2D
(b) 3D
(c) 4D
(d) 5D
ANSWER: (a)
EXPLANATION: Let the radius of curvature of the symmetric double convex lens be R each. A plane perpendicular to the principal axis that cuts the lens in equal parts, makes each part a Plano-convex lens as shown in the figure,
If the focal length of the original lens = f
1/f = (µ-1){1/R-1/(-R)} =2(µ-1)/R
Hence the power of the original lens
P = 4D =1/f = 2(µ-1)/R
→(µ-1)/R = 4D/2 =2D
For the part lens R=R and R' = - ∞, if f' = focal length, then
1/f' = (µ-1){1/R -1/(-∞)} =(µ-1)/R
Hence the power of the part lens P' =1/f'
→P' = (µ-1)/R =2D
Hence the option (a).
13. A symmetric double convex lens is cut in two equal part by a plane containing the principal axis. If the power of the original lens was 4D, the power of a cut lens will be
(a) 2D
(b) 3D
(c) 4D
(d) 5D
ANSWER: (c)
EXPLANATION: The plane containing the principal axis will cut the lens in two semicircular pieces and each piece will be a lens having the same radius of the curvatures. Hence each of the pieces will have the same focal length and the same power. Thus the option (c).
14. Two concave lenses L₁ and L₂ are kept in contact with each other. If the space between the two lenses is filled with a material of refractive index µ = 1, the magnitude of the focal length of the combination
(a) becomes undefined
(b) remains unchanged
(c) increases
(d) decreases
ANSWER: (c)
EXPLANATION: The focal length F of the combination is given as
1/F = 1/f₁ + 1/f₂
It is like resistors connected in parallel. The numerical value of equivalent resistance is less than the smallest resistor. Similarly, the numerical value of the equivalent focal length will decrease. Since the 1/F is the power of the lens and in the case of a diverging lens, it is negative as in this case. Hence F is also negative. If the numerical value of F decreases that means its negative will increase.
15. A thin lens is made with a material having a refractive index µ = 1.5. Both the sides are convex. It is dipped in water (µ = 1.33). It will behave like
(a) a convergent lens
(b) a divergent lens
(c) a rectangular slab
(d) a prism
ANSWER: (a)
EXPLANATION: The double convex lens behaves as a convergent lens in the air because the µ of the lens is greater than air. When it is dipped in water its µ is still greater than water hence it will act as a convergent lens.
16. A convex lens is made of a material having refractive index 1.2, both the surfaces of the lens are convex. If it is dipped into water (µ = 1.33). It will behave like
(a) a convergent lens
(b) a divergent lens
(c) a rectangular slab
(d) a prism
ANSWER: (b)
EXPLANATION: The µ of the convex lens is greater than air so it will act as a convergent lens in the air. When it is dipped in water its µ is less than water. Now the rays entering and exiting the lens will bend toward or away from the normal in an opposite fashion than when the lens was placed in air. So the convergent lens in the air will now behave as a divergent lens in the water.
17. A point object O is placed on the principal axis of a convex lens of focal length f = 20 cm at a distance of 40 cm to the left of it. The diameter of the lens is 10 cm. An eye is placed 60 cm to the right of the lens and a distance h below the principal axis. The maximum value of h to see the image is
(a) 0
(b) 2.5 cm
(c) 5 cm
(d) 10 cm.
ANSWER: (b)
EXPLANATION: Let us draw a diagram as below,
Since the object is placed at 2f distance, the image will also be at 2f = 40 cm distance at I. The eye is at E. AC = 5 cm, CI = 40 cm, ID = 20 cm and DE = h. Since triangles ACI and DEI are similar triangles their sides will be proportional. Hence
DE/AC = ID/CI
→h/5 = 20/40 =1/2
→h = 5/2 =2.5 cm
18. The rays of different colors fail to converge at a point after going through a converging lens. This defect is called
(a) spherical aberration
(b) distortion
(c) coma
(d) chromatic aberration
ANSWER: (d)
EXPLANATION: Since the refractive index of a medium is different for different colors. Hence the rays of different colors fail to converge at a point. Since this aberration is related to colored rays it is called a chromatic aberration because chrome is a Greek word for color.
(a) All the reflected rays meet at a point when produced backward.
(b) Only the reflected rays close to the normal meet at a point when produced backward.
(c) Only the reflected rays making a small angle with the mirror, meet at a point when produced backward.
(d) Light of different colors makes different images.
ANSWER: (a)
EXPLANATION: See the diagram below,
 |
| Diagram for Q-1 |
Take any ray PR which is reflected at Q and QR is the reflected ray. Since the angle of incidence is equal to the angle of reflection,
∠PQN =∠RQN=∠OPQ =α
RQ, when produced back, meets PO produced at P'.
In right-angled triangles PQO and P'QO,
∠QPO =∠QP'O =α
Thus both are equal in all respect.
So, PO =P'O.
For any other ray, it can be proved.
Hence the option (a).
P' becomes a virtual image of P.
2. Total internal reflection can take place only if
(a) Light goes from optically rarer medium (smaller refractive index) to optically denser medium.
(b) light goes from optically denser medium to rarer medium
(c) the refractive indices of the two media are close to each other
(d) the refractive indices of the two media are widely different.
ANSWER: (b)
EXPLANATION: When the light goes from a denser medium to lighter medium, it goes away from the normal. When the angle of incidence is increased, the angle of refraction becomes 90°. It is the critical angle. Now if the angle is increased further, the phenomenon of refraction is changed to total internal reflection.
3. In image formation from spherical mirrors, only paraxial rays are considered because they are
(a) easy to handle geometrically
(b) contain most of the intensity of the incident light
(c) form nearly a point image of a point source
(d) show a minimum dispersion effect.
ANSWER: (c)
EXPLANATION: The reflected rays from a point source in a spherical mirror do not meet at a single point. To avoid it, rays close to the principal axis (Paraxial rays) are taken. It forms nearly a point image of a point source.
4. A point object is placed at a distance of 30 cm from a convex mirror of focal length 30 cm. The image will form at
(a) infinity
(b) pole
(c) focus
(d) 15 cm behind the mirror.
ANSWER: (d)
EXPLANATION: The option can be selected by suitable reasoning. In a convex mirror, the object at infinity has its image at the focus behind the mirror. So option (c) is not correct because the object here is not at infinity.
The image in a convex mirror never forms at infinity. The option (a) is not correct.
The image will be near the pole if the object is near the pole, which is not the case here. The option (b) is not correct.
The only option left is (d) and the image position is between focal length and the pole for the object between infinity and the pole. Hence the option (d) is correct.
5. Figure (19-Q2) shows two rays A and B being reflected by a mirror and going as A' and B'. The mirror
(a) is plane
(b) is convex
(c) is concave
(d) may be any spherical mirror.
 |
| Figure for Q-5 |
ANSWER: (a)
EXPLANATION: The parallel rays after reflection are still parallel, which is possible only in a plane mirror. Concave or convex mirrors either converge or diverge the parallel rays. Hence the option (a) is correct.
6. The image formed by a concave mirror
(a) is always real
(b) is always virtual
(c) is certainly real if the object is virtual
(d) is certainly virtual if the object is real.
ANSWER: (c)
EXPLANATION: The image formed by a concave mirror may be real or virtual depending upon the position of the object. Hence the option (a) and (b) are wrong. Only the real object placed between the focus and the pole has a virtual image otherwise it is real. The option (d) is not correct.
Hence the only option remaining is (c).
7. Figure (18-Q3) shows three transparent media of refractive indices µ₁, µ₂, and µ₃. A point object O is placed in the medium µ₂. If the entire medium on the right of the spherical surface has a refractive index µ₁, the image forms at O'. If this entire medium has refractive index µ₃, the image forms at O". In the situation shown,
(a) the image forms between O' and O".
(b) the image forms to the left of O'.
(c) the image forms to the right of O".
(d) two images form, one at O' and the other at O".
 |
| Figure for Q-7 |
ANSWER: (d)
EXPLANATION: The rays incident on the surface above the principal axis will be refracted in the medium of R.I. µ₃ and form the image at O". But the rays falling below the axis will be refracted in the medium having R.I. µ₁ and the image will be formed at O'. Hence two images will be formed.
8. Four modifications are suggested in the lens formula to include the effect of the thickness t of the lens. Which one is likely to be correct?
(a) 1/v - 1/u = t/uf
(b) t/v² - 1/u = 1/f
(c) 1/(v-t) - 1/(u+t) = 1/f
(d) 1/v - 1/u +t/uv = t/f
ANSWER: (c)
EXPLANATION: The thickness of the lens will affect the object distance or image distance. Either the thickness t will be added or subtracted to u or v. It will not be in a proportion as shown in the options (a), (b) and (d). Hence the option (c) is correct.
9. A double convex lens has two surfaces of equal radii R and refractive index µ = 1.5, we have,
(a) f =R/2
(b) f =R
(c) f = -R
(d) f =2R
ANSWER: (b)
EXPLANATION: From lens makers formula
1/f = (µ-1){1/R - 1/R'}
here µ = 1.5, R' = -R
→1/f = 0.5*2/R =1/R
→f = R
Hence the option (b) is true.
10. A point source of light is placed at a distance of 2f from a converging lens of focal length f. The intensity on the other side of the lens is maximum at a distance
(a) f
(b) between f and 2f
(c) 2f
(d) more than 2f.
ANSWER: (c)
EXPLANATION: The intensity of the light on the other side is maximum where the image of the point source is formed.
Here u = -2f, f = f, hence from the lens formula
1/v -1/(-2f) = 1/f
→1/v = 1/f -1/2f = 1/2f
→v = 2f
Hence the image will be formed at 2f distance from the lens on the other side.
11. A parallel beam of light is incident on a converging lens parallel to its principal axis. As one moves away from the lens on the other side on its principal axis, the intensity of light
(a) remains constant
(b) continuously increases
(c) continuously decreases
(d) first increases then decrease.
ANSWER: (d)
EXPLANATION: The parallel beam of light converges at the focus on the other side and then diverges. Hence till the person moves to the focus the intensity will increase. Beyond the focus, it will decrease. Hence the option (d).
12. A symmetric double convex lens is cut in two equal part by a plane perpendicular to the principal axis. If the power of the original lens was 4D, the power of a cut lens will be
(a) 2D
(b) 3D
(c) 4D
(d) 5D
ANSWER: (a)
EXPLANATION: Let the radius of curvature of the symmetric double convex lens be R each. A plane perpendicular to the principal axis that cuts the lens in equal parts, makes each part a Plano-convex lens as shown in the figure,
|
| Diagram for Q-12 |
If the focal length of the original lens = f
1/f = (µ-1){1/R-1/(-R)} =2(µ-1)/R
Hence the power of the original lens
P = 4D =1/f = 2(µ-1)/R
→(µ-1)/R = 4D/2 =2D
For the part lens R=R and R' = - ∞, if f' = focal length, then
1/f' = (µ-1){1/R -1/(-∞)} =(µ-1)/R
Hence the power of the part lens P' =1/f'
→P' = (µ-1)/R =2D
Hence the option (a).
13. A symmetric double convex lens is cut in two equal part by a plane containing the principal axis. If the power of the original lens was 4D, the power of a cut lens will be
(a) 2D
(b) 3D
(c) 4D
(d) 5D
ANSWER: (c)
EXPLANATION: The plane containing the principal axis will cut the lens in two semicircular pieces and each piece will be a lens having the same radius of the curvatures. Hence each of the pieces will have the same focal length and the same power. Thus the option (c).
14. Two concave lenses L₁ and L₂ are kept in contact with each other. If the space between the two lenses is filled with a material of refractive index µ = 1, the magnitude of the focal length of the combination
(a) becomes undefined
(b) remains unchanged
(c) increases
(d) decreases
ANSWER: (c)
EXPLANATION: The focal length F of the combination is given as
1/F = 1/f₁ + 1/f₂
It is like resistors connected in parallel. The numerical value of equivalent resistance is less than the smallest resistor. Similarly, the numerical value of the equivalent focal length will decrease. Since the 1/F is the power of the lens and in the case of a diverging lens, it is negative as in this case. Hence F is also negative. If the numerical value of F decreases that means its negative will increase.
15. A thin lens is made with a material having a refractive index µ = 1.5. Both the sides are convex. It is dipped in water (µ = 1.33). It will behave like
(a) a convergent lens
(b) a divergent lens
(c) a rectangular slab
(d) a prism
ANSWER: (a)
EXPLANATION: The double convex lens behaves as a convergent lens in the air because the µ of the lens is greater than air. When it is dipped in water its µ is still greater than water hence it will act as a convergent lens.
16. A convex lens is made of a material having refractive index 1.2, both the surfaces of the lens are convex. If it is dipped into water (µ = 1.33). It will behave like
(a) a convergent lens
(b) a divergent lens
(c) a rectangular slab
(d) a prism
ANSWER: (b)
EXPLANATION: The µ of the convex lens is greater than air so it will act as a convergent lens in the air. When it is dipped in water its µ is less than water. Now the rays entering and exiting the lens will bend toward or away from the normal in an opposite fashion than when the lens was placed in air. So the convergent lens in the air will now behave as a divergent lens in the water.
17. A point object O is placed on the principal axis of a convex lens of focal length f = 20 cm at a distance of 40 cm to the left of it. The diameter of the lens is 10 cm. An eye is placed 60 cm to the right of the lens and a distance h below the principal axis. The maximum value of h to see the image is
(a) 0
(b) 2.5 cm
(c) 5 cm
(d) 10 cm.
ANSWER: (b)
EXPLANATION: Let us draw a diagram as below,
 |
| Diagram for Q-17 |
Since the object is placed at 2f distance, the image will also be at 2f = 40 cm distance at I. The eye is at E. AC = 5 cm, CI = 40 cm, ID = 20 cm and DE = h. Since triangles ACI and DEI are similar triangles their sides will be proportional. Hence
DE/AC = ID/CI
→h/5 = 20/40 =1/2
→h = 5/2 =2.5 cm
18. The rays of different colors fail to converge at a point after going through a converging lens. This defect is called
(a) spherical aberration
(b) distortion
(c) coma
(d) chromatic aberration
ANSWER: (d)
EXPLANATION: Since the refractive index of a medium is different for different colors. Hence the rays of different colors fail to converge at a point. Since this aberration is related to colored rays it is called a chromatic aberration because chrome is a Greek word for color.
===<<<O>>>===
My Channel on YouTube → SimplePhysics with KK
Links to the Chapters
Links to the Chapters
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 18 - Geometrical Optics
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
EXERCISES -Q-51 TO Q-60
EXERCISES -Q-61 TO Q-70
EXERCISES - Q-71 TO Q-80
EXERCISES - Q-81 TO Q-89
CHAPTER- 15 - Wave Motion and Waves on a String
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Question for Short Answers
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → Exercises (1-10)
Click here for → Exercises (11-20)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
CHAPTER- 7 - Circular Motion
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
Click here for → Questions for Short Answer
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these- "New Questions on Friction".
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
---------------------------------------------------------------------------------
---------------------------------------------------------------------------------
CHAPTER- 5 - Newton's Laws of Motion
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
-------------------------------------------------------------------------------
-------------------------------------------------------------------------------
CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
--------------------------------------------------------------------------------------------------------------
--------------------------------------------------------------------------------------------------------------
CHAPTER- 3 - Kinematics - Rest and Motion
Click here for "Questions for short Answers"
Click here for "OBJECTIVE-I"
Click here for EXERCISES (Question number 1 to 10)
Click here for EXERCISES (Question number 11 to 20)
Click here for EXERCISES (Question number 21 to 30)
Click here for EXERCISES (Question number 31 to 40)
Click here for EXERCISES (Question number 41 to 52)
CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
No comments:
Post a Comment