Chapter 24, Problem 16Q

(a)

To determine

To explain: Change in the diffraction pattern of a single slit if the whole apparatus is immersed in water.

Answer to Problem 16Q

The diffraction pattern gets smaller.

Explanation of Solution

Introduction: The pattern resulting from superimposing of two waves is known as interference. Interference can be destructive or constructive.

Destructive interference occurs after the crest of a wave overlaps the trough of another wave. The waves destructively interfere and also, cancel each other, once crest of one & trough of other interfere. To find destructive interference for single slit is

  $a\sin \theta =m\lambda$

Where,

  $m$ is an integer

  $\theta$ is the angle relative to original direction of light,

  $\lambda$ is wavelength

  $a$ is a slit width

When a student immerses single-slit diffraction apparatus in water, due to increase in index of refraction as well as diffraction pattern gets more closely spaced, wavelength of the light gets smaller.

(b)

To determine

To explain: The change in the diffraction pattern of a single slit if the whole apparatus is immersed in vacuum.

Answer to Problem 16Q

The diffraction pattern gets lightly larger.

Explanation of Solution

Introduction: The pattern resulting from superimposing of two waves is known as interference. Interference can be destructive or constructive.

Destructive interference occurs after the crest of a wave overlaps the trough of another wave. The waves destructively interfere and also, cancel each other, once crest of one & trough of other interfere. To find destructive interference for single slit is

  $a\sin \theta =m\lambda$

Where,

  $m$ is an integer

  $\theta$ is the angle relative to original direction of light,

  $\lambda$ is wavelength

  $a$ is a slit width

When a student immerses single-slit diffraction apparatus in vacuum, due to decrease in index of refraction as well as diffraction pattern spreads farther apart, wavelength of the light gets larger.