Wave Optics Easy Questions and Answers | Page - 2

Wave Optics

In the Young’s double-slit experiment, the intensity of light at a point on the screen where the path difference is λ is K, (λ being the wave length of light used). The intensity at a point where the path difference is λ /4, will be:

1. K
1. K/4
1. K/2
1. Zero

View Hint View Answer Discuss in Forum

For path difference λ, phase difference = 2π rad. For path difference λ/4, phase difference
= π rad.
2

As K = 4I₀ so intensity at given point where path difference is
λ
4

K' = 4I₀ cos² π cos π = cos45°
4 4

= 2I₀ = K
2

Correct Option: C

For path difference λ, phase difference = 2π rad. For path difference λ/4, phase difference
= π rad.
2

As K = 4I₀ so intensity at given point where path difference is
λ
4

K' = 4I₀ cos² π cos π = cos45°
4 4

= 2I₀ = K
2

Two slits in Young’s experiment have widths in the ratio 1 : 25. The ratio of intensity at the maxima and minima in the interference pattern,
I_max is :
I_min

1. 121
  49
1. 49
  121
1. 4
  9
1. 9
  4

View Hint View Answer Discuss in Forum

The ratio of slits width
= 1 (given)
25

∴ I₁ = 25
I₂ 1

I ∝ A² ⇒ I₁ = A₁² = 25
I₂ A₂² 1

or A₁ = 5
A₂ 1

A_max = A₁ + A₂
A_min A₁ - A₂

= 5 + 1 = 6 = 3
5 - 1 4 2

= 3 ²
2

= 9
4

Correct Option: D

The ratio of slits width
= 1 (given)
25

∴ I₁ = 25
I₂ 1

I ∝ A² ⇒ I₁ = A₁² = 25
I₂ A₂² 1

or A₁ = 5
A₂ 1

A_max = A₁ + A₂
A_min A₁ - A₂

= 5 + 1 = 6 = 3
5 - 1 4 2

= 3 ²
2

= 9
4

In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light wavelength
500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern ?

1. 0.1 mm
1. 0.5 mm
1. 0.02 mm
1. 0.2 mm

View Hint View Answer Discuss in Forum

Here, distance between two slits,
d = 1 mm = 10^–3 m
distance of screen from slits, D = 1 m
wavelength of monochromatic light used,
λ = 500 nm = 500 × 10^–9 m
width of each slit a = ?
Width of central maxima in single slit pattern
= 2λ D
a

Fringe width in double slit experiment
β = λ D
d

So,required condition
10λ D = 2λ D
d a

⇒ a = d = 1 × 10^-3 m = 0.2 mm
5D 5

Correct Option: D

Here, distance between two slits,
d = 1 mm = 10^–3 m
distance of screen from slits, D = 1 m
wavelength of monochromatic light used,
λ = 500 nm = 500 × 10^–9 m
width of each slit a = ?
Width of central maxima in single slit pattern
= 2λ D
a

Fringe width in double slit experiment
β = λ D
d

So,required condition
10λ D = 2λ D
d a

⇒ a = d = 1 × 10^-3 m = 0.2 mm
5D 5

Colours appear on a thin soap film and on soap bubbles due to the phenomenon of

1. refraction
1. dispersion
1. interference
1. diffraction

View Hint View Answer Discuss in Forum

We know that the colours for which the condition of constructive interference is satisfied are observed in a given region of the film. The path difference between the light waves reaching the eye changes when the position of the eye is changed. Therefore, colours appear on a thin soap film or soap bubbles due to the phenomenon of interference.

Correct Option: C

We know that the colours for which the condition of constructive interference is satisfied are observed in a given region of the film. The path difference between the light waves reaching the eye changes when the position of the eye is changed. Therefore, colours appear on a thin soap film or soap bubbles due to the phenomenon of interference.

Interference was observed in interference chamber where air was present, now the chamber is evacuated, and if the same light is used, a careful observer will see

1. no interference
1. interference with brighter bands
1. interference with dark bands
1. interference fringe with larger width

View Hint View Answer Discuss in Forum

In vaccum, λ increases very slightly compared to that in air. As β ∝ λ, therefore, width of interference fringe increases slightly.

Correct Option: D

In vaccum, λ increases very slightly compared to that in air. As β ∝ λ, therefore, width of interference fringe increases slightly.