Q#11
The equation of refraction at a spherical surface isµ₂/v - µ₁/u = (µ₂ - µ₁)/R
Taking R = ∞, show that the equation leads to the equation
Real depth/Apparent depth = µ₁/µ₂
for refraction at a plane surface.
Answer:
The equation of refraction at a spherical surface is
µ₂/v - µ₁/u = (µ₂ - µ₁)/R
Given that R = ∞, (Implies that the surface is plane)
Hence, µ₂/v - µ₁/u = 0
µ₂/v = µ₁/u
µ₁/µ₂ = u/v = h₁/h₂ = Real Depth/Apparent depth
Q#12
A thin converging lens is formed with one surface convex and the other plane. Does the position of the image depend on whether the convex surface or plane surface faces the object?
Answer:
From the lens makers formula
1/f = (µ - 1){1/R₁ - 1/R₂}
Case-I, when the object is facing the convex surface.
R₂ = ∞, and R₁ = R₁
1/f = (µ - 1){1/R₁ - 0} = (µ - 1)/R₁
Case-II, when the object is facing the plane surface of the lens.
R₁ = ∞, and R₂ = -R₁ {since the center of curvature is towards left of origin at P}
1/f = (µ - 1){0 - 1/(-R₁)} = (µ - 1)/R₁
Thus in both cases, the focal length does not change and the position of the image does not change whether the convex surface or the plane surface faces the object.
Q#13
A single lens is mounted in a tube. A parallel beam enters the tube and emerges out of the tube as a divergent beam. Can you say with certainty that there is a diverging lens in the tube?
Answer:
Assuming that the tube is in the air and the lens is made of a material having the refractive index µ > 1, it can not be said with certainty that there is a diverging lens (concave lens) inside the tube. A converging lens (convex lens) having its focal point well inside the tube will also let this beam emerge out as a diverging beam.
Q#14
An air bubble is formed inside water. Does it act as a converging lens or a diverging lens?
Answer:
Though the air bubble inside the water has convex surfaces, it acts as a diverging lens because in this case the refractive index of the lens (air bubble) is less than the surrounding. Hence it will act just opposite than the usual glass lens in the air which acts as a converging lens.
Q#15
Two converging lenses of unequal focal lengths can be used to reduce the aperture of a parallel beam of light without losing the energy of light. This increases the intensity. Describe how the converging lenses should be placed to do this.
Answer:
It can be achieved by placing the larger focal length lens towards the incoming beam and then the smaller focal length lens at a distance f+f' from the first lens. See the diagram below.
In the above diagram, the focal length of the lens AB = f and the focal length of the lens CD = f', such that
f > f'.
The aperture of the incoming parallel beam is AB. The first lens converges it at the point F, PF = f. Since F is also the focal point of the lens CD, the rays coming from F become a parallel beam of aperture CD after passing through the second lens. The energy of the beam remains the same after passing through the two lenses. The triangle APF and FP'D are similar triangles hence
PF/P'F = AP/DP' = 2AP/2DP' = AB/CD
AB/CD = PF/P'F = f/f'
Since f > f', AB > CD
So the aperture is reduced by this arrangement. The intensity is the energy per unit area. The energy of the light beam is constant but aperture (hence the area) reduces, thus intensity increases in this process.
Q#16
If a spherical mirror is dipped in water, does its focal length change?
Answer:
The mirrors reflect rays and incident and reflected rays remain in the same medium, its focal length f = R/2 does not depend on the refractive index of the medium in which it is placed. So if a spherical mirror is dipped in water its focal length will not change.
Q#17
If a thin lens is dipped in water, does its focal length change?
Answer:
The focal length of a lens is given as
1/f = (µ₂/µ₁ - 1){1/R₁ -1/R₂}
When dipped in water, the factor µ₂/µ₁ changes. Hence the focal length will also change.
Q#18
Can mirrors give rise to chromatic aberration?
Answer:
In mirrors laws of reflection are followed. The angle of incidence and the angle of reflection are the same for each wavelength of the light. Hence the focal length of a mirror is the same for each wavelength of the light. Thus mirrors can not give rise to chromatic aberration.
Q#19
Laser light is focused by a converging lens. Will there be a significant chromatic aberration?
Answer:
Laser lights are monochromatic i.e. its wavelengths are nearly the same. Thus they will focus at the same point and no significant chromatic
R₁ = ∞, and R₂ = -R₁ {since the center of curvature is towards left of origin at P}
1/f = (µ - 1){0 - 1/(-R₁)} = (µ - 1)/R₁
Thus in both cases, the focal length does not change and the position of the image does not change whether the convex surface or the plane surface faces the object.
Q#13
A single lens is mounted in a tube. A parallel beam enters the tube and emerges out of the tube as a divergent beam. Can you say with certainty that there is a diverging lens in the tube?
Answer:
Assuming that the tube is in the air and the lens is made of a material having the refractive index µ > 1, it can not be said with certainty that there is a diverging lens (concave lens) inside the tube. A converging lens (convex lens) having its focal point well inside the tube will also let this beam emerge out as a diverging beam.
Q#14
An air bubble is formed inside water. Does it act as a converging lens or a diverging lens?
Answer:
Though the air bubble inside the water has convex surfaces, it acts as a diverging lens because in this case the refractive index of the lens (air bubble) is less than the surrounding. Hence it will act just opposite than the usual glass lens in the air which acts as a converging lens.
Q#15
Two converging lenses of unequal focal lengths can be used to reduce the aperture of a parallel beam of light without losing the energy of light. This increases the intensity. Describe how the converging lenses should be placed to do this.
Answer:
It can be achieved by placing the larger focal length lens towards the incoming beam and then the smaller focal length lens at a distance f+f' from the first lens. See the diagram below.
f > f'.
The aperture of the incoming parallel beam is AB. The first lens converges it at the point F, PF = f. Since F is also the focal point of the lens CD, the rays coming from F become a parallel beam of aperture CD after passing through the second lens. The energy of the beam remains the same after passing through the two lenses. The triangle APF and FP'D are similar triangles hence
PF/P'F = AP/DP' = 2AP/2DP' = AB/CD
AB/CD = PF/P'F = f/f'
Since f > f', AB > CD
So the aperture is reduced by this arrangement. The intensity is the energy per unit area. The energy of the light beam is constant but aperture (hence the area) reduces, thus intensity increases in this process.
Q#16
If a spherical mirror is dipped in water, does its focal length change?
Answer:
The mirrors reflect rays and incident and reflected rays remain in the same medium, its focal length f = R/2 does not depend on the refractive index of the medium in which it is placed. So if a spherical mirror is dipped in water its focal length will not change.
Q#17
If a thin lens is dipped in water, does its focal length change?
Answer:
The focal length of a lens is given as
1/f = (µ₂/µ₁ - 1){1/R₁ -1/R₂}
When dipped in water, the factor µ₂/µ₁ changes. Hence the focal length will also change.
Q#18
Can mirrors give rise to chromatic aberration?
Answer:
In mirrors laws of reflection are followed. The angle of incidence and the angle of reflection are the same for each wavelength of the light. Hence the focal length of a mirror is the same for each wavelength of the light. Thus mirrors can not give rise to chromatic aberration.
Q#19
Laser light is focused by a converging lens. Will there be a significant chromatic aberration?
Answer:
Laser lights are monochromatic i.e. its wavelengths are nearly the same. Thus they will focus at the same point and no significant chromatic
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