Questions OBJECTIVE - I and Answer (Geometrical Optics) HC Verma Part 1 (10-18)

 Q#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)

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.

Q#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)
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).

Q#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)
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).

Q#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)
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).

Q#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)
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.

Q#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)
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.

Q#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)
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.

Q#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)
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

Q#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)
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.

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