Question

In Young's double slit interference experiment, using two coherent waves of different amplitudes, the intensity ratio between bright and dark fringes is 3. Then, the value of the ratio of the amplitudes of the waves that arrive there is

• A. \( \frac{\sqrt{3}+1}{\sqrt{3}-1} \)
• B. \( \frac{\sqrt{3}}{1} \)
• C. \( 3:1 \)
• D. \( 1: \sqrt{3} \)

Hint:

In Young's double slit experiment, the intensity ratio between bright and dark fringes can help determine the amplitude ratio between the waves. The formula \( \frac{I_{\text{bright}}}{I_{\text{dark}}} = 3 \) is key to solving for the amplitude ratio.

Answer 1

The Correct Option is A

In Young's double slit experiment, the intensity at any point on the screen is given by: \[ I = I_0 \left( 1 + \cos \delta \right) \] where \( I_0 \) is the maximum intensity and \( \delta \) is the phase difference between the two waves. 
For bright and dark fringes, the intensity can be written in terms of the amplitudes \( A_1 \) and \( A_2 \) of the two waves as follows: 

- For bright fringes: \( I_{\text{bright}} = I_0 \left(1 + \frac{A_1}{A_2}\right)^2 \) 
- For dark fringes: \( I_{\text{dark}} = I_0 \left(1 - \frac{A_1}{A_2}\right)^2 \) 
Given that the intensity ratio between bright and dark fringes is 3, we can equate the intensities: \[ \frac{I_{\text{bright}}}{I_{\text{dark}}} = 3 \] Substituting the expressions for \( I_{\text{bright}} \) and \( I_{\text{dark}} \): \[ \frac{\left(1 + \frac{A_1}{A_2}\right)^2}{\left(1 - \frac{A_1}{A_2}\right)^2} = 3 \] Taking square roots on both sides: \[ \frac{1 + \frac{A_1}{A_2}}{1 - \frac{A_1}{A_2}} = \sqrt{3} \] Solving for \( \frac{A_1}{A_2} \): \[ 1 + \frac{A_1}{A_2} = \sqrt{3} \left( 1 - \frac{A_1}{A_2} \right) \] Expanding: \[ 1 + \frac{A_1}{A_2} = \sqrt{3} - \sqrt{3} \cdot \frac{A_1}{A_2} \] Rearranging terms: \[ \frac{A_1}{A_2} \left( 1 + \sqrt{3} \right) = \sqrt{3} - 1 \] Thus: \[ \frac{A_1}{A_2} = \frac{\sqrt{3}-1}{\sqrt{3}+1} \] 
Hence, the ratio of the amplitudes of the two waves is \( \frac{\sqrt{3}+1}{\sqrt{3}-1} \). 
Thus, the correct answer is: \[ \text{(A) } \frac{\sqrt{3}+1}{\sqrt{3}-1} \]

Answer 2

To solve the problem, we first need to understand the relationship between intensity and amplitude in wave interference. In Young's double-slit experiment, the intensity at a point is proportional to the square of the amplitude of the resultant wave. For two coherent waves with amplitudes \(a_1\) and \(a_2\), the resultant amplitude \(A\) at a bright fringe due to constructive interference is given by:

\(A_{\text{bright}}=a_1+a_2\)

and at a dark fringe due to destructive interference, it is:

\(A_{\text{dark}}=|a_1-a_2|\)

The intensity \(I\) is proportional to the square of these amplitudes:

\(I_{\text{bright}}=(a_1+a_2)^2\)

\(I_{\text{dark}}=(a_1-a_2)^2\)

We are given that the intensity ratio of bright to dark fringes is \(3\):

\(\frac{I_{\text{bright}}}{I_{\text{dark}}}=3\)

Substituting in the expressions for intensity, we get:

\(\frac{(a_1+a_2)^2}{(a_1-a_2)^2}=3\)

Taking the square root on both sides gives:

\(\frac{a_1+a_2}{a_1-a_2}=\sqrt{3}\)

Solving for the ratio \(\frac{a_1}{a_2}\), we let \(x=\frac{a_1}{a_2}\). Then:

\(\frac{x+1}{x-1}=\sqrt{3}\)

Cross-multiplying gives:

\(x+1=\sqrt{3}(x-1)\)

\(x+1=\sqrt{3}x-\sqrt{3}\)

Rearranging terms, we have:

\(x-\sqrt{3}x=-\sqrt{3}-1\)

\(x(1-\sqrt{3})=-\sqrt{3}-1\)

Solving for \(x\), we obtain:

\(x=\frac{\sqrt{3}+1}{\sqrt{3}-1}\)

Thus, the ratio of the amplitudes of the waves is:

\(\boxed{\frac{\sqrt{3}+1}{\sqrt{3}-1}}\)