Wave Optics Test

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Wave Optics Test - 4

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Question 1/10
1 / -0

From a medium of index of refraction n₁, monochromatic light of wavelength λ is incident normally on a thin film of uniform thickness L ( L > 0.1 λ ) and refractive index n₂. The light transmitted by the film travels into a medium with refractive index n₃. Minimum film thickness for which maximum light is transmitted is ( given n₁ < n₂ < n₃ )

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A

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B

Correct

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C

Correct

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D

Solutions

Path difference for maxima in transmission is given by
Question 2/10
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A plastic sheet of refractive index = 1.6 covers one slit of a double slit arrangement. When the double slit is illuminated by monochromatic light (wavelength in air = 6600 Å), the centre of the screen appears dark rather than bright. The minimum thickness of the plastic sheet to be used for this to happen is

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A
5500 Å

Correct

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B
6600 Å

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C
3300 Å

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D
2200 Å

Solutions

The path difference produced by a sheet

Δ = ( μ - 1 ) t

For minimum thickness
Question 3/10
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For a Young's double slit experiment shown in figure, given d << ?, d << D, λ/D << 1. Which of the following is a right statement about the wavelength of light used?

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A
The central maxima of all wavelengths will coincide

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B
If white light is used, violet color forms its first maxima closest to the central maxima.

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C
All of the above

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D
Larger the wavelength, larger will be the fringe width.

Solutions

⇒ b ∝ λ (fringe width is proportional to wavelength )

However, position of central maxima is independent of wavelength used

⇒ y_n ∝ λ

Maxima of waves with smaller wavelengths will be closer to the central maximum.
Wavelength increases from Violet towards Red
Question 4/10
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Consider the YDSE arrangement shown in figure. If d = 10 λ, then position of 8^th maxima is

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A

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B

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C

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D

Solutions

For maxima
d sin θ = n λ
Question 5/10
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A Young's double slit experiment is conducted in a liquid of refractive index μ₁ . A glass plate of thickness ‘t’ and refractive index μ₂ is placed in the path of one slit. The magnitude of the optical path difference at centre of screen will be

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A

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B
|(μ₂ – μ₁) t|

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C
|(μ₂ – 1) t|

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D

Solutions

μ₁ (S₁O) - μ₁ (S₂O) + μ₁ t - μ₂ t = 0

μ₁ ( S₁O - S₂O ) = ( μ₂ - μ₁ ) t
Question 6/10
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In an experiment to find the diameter of a human hair, the hair is placed between two flat glass plates, and the setup is illuminated with light of wavelength λ = 552 nm ( in vaccum ). The number of bright fringes produced between the edge of the plates and the hair are found to be 125. What is the diameter of the hair?

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A
5.25 × 10^–8 m

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B
3.44 × 10^–5 m

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C
4.66 × 10^–7 m

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D
2.55 × 10^–6 m

Solutions

Reflections from the boundaries will cause a net phase shift of 180°
The condition for bright fringes is

2 t = (m + 1/2) λ_film

Now, m = 124 since there is a bright fringe for m = 0 and

= 3.44 × 10^-5 m
Question 7/10
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The sky is blue because

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A
dust particles in the air are bluish in color

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B
air molecules absorb red light more efficiently than blue light because of their orbital electrons

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C
tiny particles in the air are more efficient at scattering light of short wavelength than light of long wavelength

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D
air molecules absorb blue light more efficiently than red light because of their orbital electrons

Solutions

We perceive the color of sky from the light scattered by our atmosphere.
Question 8/10
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A certain region of a soap bubble reflects red light of vaccum wavelength λ = 650 nm. What is the minimum thickness that this region of the soap bubble could have? Take the index of reflection of the soap film to be 1.41

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A
2.4 × 10^–7 m

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B
240 × 10⁷ m

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C
120 × 10⁷ m

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D
1.2 × 10^–7 m

Solutions

There is air on both sides of the soap film.

∴ the reflections of the light produce a net phase shift of 180° .

Condition for bright fringes is

2 t = (m + 1/2) λ_film
Question 9/10
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A double slit arrangement produces fringes for λ = 5890 Å that are 0.4° apart. What is the angular width if the entire arrangement is immersed in water ?
( given μ_w = 4/3 )

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A
0.9°

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B
1.5°

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C
0.3°

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D
2.4°

Solutions

Let θ be the angular width in water.

We know angular width =

⇒ Angular width ∝ λ
Question 10/10
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A ray of light travels through a slab as shown.

The refractive index of the material of the slab varies as

What is the equivalent optical path of the glass slab ?

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A
1.36 m

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B
2.54 m

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C
0.15 m

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D
3.35 m

Solutions

Equivalent optical path length will be

≈ 1.36 m

Question 1/10

Question 2/10

5500 Å

6600 Å

3300 Å

2200 Å

Question 3/10

The central maxima of all wavelengths will coincide

If white light is used, violet color forms its first maxima closest to the central maxima.

All of the above

Larger the wavelength, larger will be the fringe width.

Question 4/10

Question 5/10

|(μ₂ – μ₁) t|

|(μ₂ – 1) t|

Question 6/10

5.25 × 10^–8 m

3.44 × 10^–5 m

4.66 × 10^–7 m

2.55 × 10^–6 m

Question 7/10

dust particles in the air are bluish in color

air molecules absorb red light more efficiently than blue light because of their orbital electrons

tiny particles in the air are more efficient at scattering light of short wavelength than light of long wavelength

air molecules absorb blue light more efficiently than red light because of their orbital electrons

Question 8/10

2.4 × 10^–7 m

240 × 10⁷ m

120 × 10⁷ m

1.2 × 10^–7 m

Question 9/10

0.9°

1.5°

0.3°

2.4°

Question 10/10

1.36 m

2.54 m

0.15 m

3.35 m

Question 1/10

Question 2/10

5500 Å

6600 Å

3300 Å

2200 Å

Question 3/10

The central maxima of all wavelengths will coincide

If white light is used, violet color forms its first maxima closest to the central maxima.

All of the above

Larger the wavelength, larger will be the fringe width.

Question 4/10

Question 5/10

|(μ2 – μ1) t|

|(μ2 – 1) t|

Question 6/10

5.25 × 10–8 m

3.44 × 10–5 m

4.66 × 10–7 m

2.55 × 10–6 m

Question 7/10

dust particles in the air are bluish in color

air molecules absorb red light more efficiently than blue light because of their orbital electrons

tiny particles in the air are more efficient at scattering light of short wavelength than light of long wavelength

air molecules absorb blue light more efficiently than red light because of their orbital electrons

Question 8/10

2.4 × 10–7 m

240 × 107 m

120 × 107 m

1.2 × 10–7 m

Question 9/10

0.9°

1.5°

0.3°

2.4°

Question 10/10

1.36 m

2.54 m

0.15 m

3.35 m

|(μ₂ – μ₁) t|

|(μ₂ – 1) t|

5.25 × 10^–8 m

3.44 × 10^–5 m

4.66 × 10^–7 m

2.55 × 10^–6 m

2.4 × 10^–7 m

240 × 10⁷ m

120 × 10⁷ m

1.2 × 10^–7 m