Questions OBJECTIVE - I and Answer (Light Waves) HC Verma Part 1 (10-17)

 Q#10

The wavefronts of light coming from a distant source of unknown shape are nearly
(a) plane
(b) elliptical
(c) cylindrical
(d) spherical.

Answer:  (a)
The wavefronts of light coming from a distant source are nearly plane because at a large distance the curve is magnified and a small portion of it is nearly plane.

Q#11
The inverse square law of intensity (i.e. intensity ∝ 1/r²) is valid for a
(a) point source
(b) line source
(c) plane source
(d) cylindrical source.

Answer:  (a)
Only for a point source the intensity can be written as I = E/4ฯ€r², where E is the strength of the source. Hence
E ∝ 1/r².

Q#12
Two sources are called coherent if they produce waves
(a) of equal wavelength
(b) of equal velocity
(c) having same shape of wavefront
(d) having a constant phase difference.

Answer::  (d)
By definition, the two sources are called coherent if the initial phase difference does not change with time i.e. the phase difference remains constant.

Q#13
When a drop of oil is spread on a water surface, it displays beautiful colors in daylight because of
(a) dispersion of light
(b) reflection of light
(c) polarization of light
(d) interference of light

Answer::  (d)
The drop of oil spreads on the water surface and makes a thin film. When the daylight is incident on the upper surface of the film, a part is reflected in the air while another is transmitted and falls on the inner surface. Here again, the light is reflected and transmitted in parts. The reflected light goes through this process multiple times and these multiple reflected rays have path differences and interfere with each other. If the following condition fulfilled for a certain wavelength, then this color is illuminated and display beautiful colors. 

2ยตd = (n + ½)๐œ†

where d = thickness of the film and ๐œ† = wavelength of a certain color.

Q#14
Two coherent sources of different intensities send waves which interfere. The ratio of maximum intensity to the minimum intensity is 25. The intensities of the sources are in the ratio
(a) 25 : 1
(b) 5 : 1
(c) 9 : 4
(d) 625 : 1

Answer::  (c)
Since the resultant field at a point is given as

E₀² = E₁² + E₂² + 2E₁E₂ cos๐›ฟ

Where E₁ and E₂ = amplitudes of interfering waves and ๐›ฟ is phase difference. For maximum amplitude cos๐›ฟ = 1 and for the minimum cos๐›ฟ = -1, hence maximum amplitude is given as

E₀² = (E₁ + E₂)²

and the minimum amplitude as

E₀'² = (E₁ - E₂)²

Since the intensity is proportional to the square of the amplitude, hence the maximum intensity

I = (√I₁ + √I₂)²

and the minimum intensity

I' = (√I₁ - √I₂)²
Let I₁ = k I₂

Hence, I = (√k+1)² I₂
and I' = (√k - 1)² I₂
Given, I/I' = 25

(√k+1)²/(√k - 1)² = 25
√k + 1 = 5√k - 5
4√k = 6
√k = 6/4 = 3/2
k = 9/4

Hence the intensities of the sources are in the ratio 9:4.

Q#15
The slits in a Young's double slit experiment have equal width and the source is placed symmetrically with respect to the slits. The intensity at the central fringe is I₀. If one of the slits is closed, the intensity at this point will be
(a) I₀
(b) I₀/4
(c) I₀/2
(d) 4I₀

Answer:  (b)
As in the previous explanation the maximum intensity

I₀ = (√I₁ + √I₂)², here given that I₁ = I₂ = I (say).

I₀ = 4I  ------- (i)

If one slit is closed I₀' = I

I₀' = I₀/4      {from (i)}
Hence option (b)

Q#16
A thin transparent sheet is placed in front of a Young's double slit. The fringe width will
(a) increase
(b) decrease
(c) remain same
(d) become non-uniform

Answer:  (c)
The fringe width is given as

w = D๐œ†/d

where D = distance of the screen from the slits
d = distance of slits.

The wavelength ๐œ† remains the same when the light comes out of the film. Thus none of the entity D, d or ๐œ† changes. So, the fringe with will remain the same. Hence the option (c).

Q#17
If Young's double slit experiment is performed in water,
(a) the fringe width will decrease
(b) the fringe width will increase
(c) the fringe width will remain unchanged
(d) there will be no fringe

Answer: (a)
As the wavelength of light in water will decrease, hence the fringe width will decrease because the fringe width is proportional to the wavelength.

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