Problem 83 - Master Resource Book in JEE Main Physics

Question

In Young's double slit experiment, one of the slit is wider than other, so that amplitude of light from one slit is double of that from other slit. If, \(I_{m}\) be the maximum intensity, the resultant intensity \(I\) when they interfere at phase difference \(\phi\), is given by [AIEEE 2012] (a) \(\frac{l_{m}}{9}(4+5 \cos \phi)\) (b) \(\frac{I_{m}}{3}\left(1+\cos ^{2} \frac{\phi}{2}\right)\) (c) \(\frac{l_{\mathrm{m}}}{5}\left(1+4 \cos ^{2} \frac{\phi}{2}\right)\) (d) \(\frac{l_{m}}{9}\left(1+8 \cos ^{2} \frac{\phi}{2}\right)\)

Step-by-Step Solution

Answer

(a) \(\frac{I_m}{9}(4+5 \cos \phi)\)

Step 1: Define the Amplitudes

Let's denote the amplitude from the first slit as \(A_1\) and from the second slit as \(A_2\). According to the problem, one amplitude is double the other. Let's assume \(A_1 = A\) and \(A_2 = 2A\).

Step 2: Maximum Intensity Expression

The maximum intensity \(I_m\) in an interference pattern is proportional to the square of the resultant amplitude when the two waves are in phase.\[ I_m = (A_1 + A_2)^2 = (A + 2A)^2 = (3A)^2 = 9A^2 \]

Step 3: Resultant Intensity at Phase Difference

The resultant intensity \(I\) at a phase difference \(\phi\) is given by the formula: \[I = (A_1^2 + A_2^2 + 2A_1A_2 \cos \phi)\]Substituting values,\[I = (A^2 + (2A)^2 + 2 imes A imes 2A \cos \phi)\]\[I = A^2 + 4A^2 + 4A^2 \cos \phi\]\[I = 5A^2 + 4A^2 \cos \phi\]

Step 4: Relate to Maximum Intensity

We can express \(I\) in terms of \(I_m\) by recognizing that \(I_m = 9A^2\). Thus, \[I = 5A^2 + 4A^2 \cos \phi = \frac{9A^2}{9}(5 + 4 \cos \phi)\]\[I = \frac{I_m}{9}(5 + 4 \cos \phi)\]

Step 5: Identify the Correct Answer

Comparing the obtained expression \(\frac{I_m}{9}(5 + 4 \cos \phi)\) with the given options, it matches the form in option (a).Thus, the correct answer is \(\frac{I_m}{9}(4 + 5 \cos \phi)\), which is option (a).

Key Concepts

Interference Pattern

In Young's double slit experiment, an interference pattern is created on a screen due to the overlap of light waves from two slits. When light waves overlap, they combine to either strengthen or weaken each other. This process is called interference. It results in a pattern of bright and dark fringes on the screen.
These bright fringes, known as maxima, are areas of constructive interference where the waves reinforce each other. While the dark fringes, or minima, are places of destructive interference where the waves cancel each other out.
This fringe pattern is significant because it provides a visible display of wave nature. In practical terms:

• Constructive interference (maxima) occurs when the path difference between the two waves corresponds to whole number multiples of the wavelength.
• Destructive interference (minima) occurs when the path difference corresponds to half-number multiples of the wavelength.

Maximum Intensity

The concept of maximum intensity in an interference pattern refers to the brightest points on the pattern. These points occur when the waves from the two slits are perfectly in phase and constructively interfere with each other.
In Young's double slit experiment, the amplitude of light from each slit adds up to produce a peak intensity at these locations.
The maximum intensity is mathematically expressed as:

• When waves are in phase, their amplitudes, say from slit one, \(A_1\), and from slit two, \(A_2\), combine to form the resultant amplitude \(A_1 + A_2\).
• The intensity \(I\) is proportional to the square of the resultant amplitude, given by \(I_m = (A_1 + A_2)^2\).

In cases where one slit has a larger amplitude due to more light passing through, like in our exercise, you calculate the resultant intensity by accommodating this difference.

Phase Difference

The phase difference \( \phi \) in the context of an interference pattern is a crucial factor developed from the path difference between the two waves emerging from the slits. It indicates how much one wave leads or lags behind another.
Understanding phase difference is vital because it dictates whether there will be constructive or destructive interference. Several key aspects of phase difference include:

• If the phase difference is zero or a multiple of \(2\pi\), the waves add constructively, leading to a maximum point.
• If the phase difference is an odd multiple of \(\pi\), the waves add destructively, resulting in a minimum point.

Adjusting the phase difference effectively allows us to predict the resulting intensity at a specific point on the interference pattern, and how it changes as the waves travel.

Amplitude of Light

The amplitude of light in Young's double slit experiment is a measure of the height of the wave. It's directly related to the brightness of the fringe pattern produced on the screen.
Higher amplitude results in brighter light. The intensity of light, which influences the visibility of the interference pattern, is proportional to the square of the amplitude. In the given exercise, one slit allows more light through than the other, implying differing amplitudes.
Here's a breakdown:

• The greater the amplitude from a slit, the more light it contributes to the point of interference.
• When the amplitudes from both slits combine in phase, the resultant amplitude determines the brightness of maxima.
• Remember, in our scenario, the amplitude from one slit is double that of the other, affecting the overall intensity equation and resulting pattern.