Question

Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :

• A. \( (2\sqrt{2} + 1) : (2\sqrt{2} - 1) \)
• B. \( (3 + 2\sqrt{2}) : (3 - 2\sqrt{2}) \)
• C. \( 9 : 1 \)
• D. \( 3 : 1 \)

Hint:

In Young's double slit experiment, the intensity of light is proportional to the width of the slit. The maximum intensity is \( (\sqrt{I_1} + \sqrt{I_2})^2 \) and the minimum intensity is \( (\sqrt{I_1} - \sqrt{I_2})^2 \), where \( I_1 \) and \( I_2 \) are the intensities from the two slits. Use the given relationship between the widths to find the ratio of intensities and then calculate the ratio of maximum to minimum intensity.

Answer 1

The Correct Option is B

In Young's double slit experiment, the intensity of light passing through a slit is directly proportional to the width of the slit. 
Let the widths of the two slits be \( w_1 \) and \( w_2 \). Given that the width of one slit is half the width of the other slit, let \( w_1 = w \) and \( w_2 = 2w \). 
The intensities of light from the two slits are proportional to their widths. Let the intensities be \( I_1 \) and \( I_2 \). \[ I_1 \propto w_1 = w \implies I_1 = I_0 \] \[ I_2 \propto w_2 = 2w \implies I_2 = 2I_0 \] The maximum intensity \( I_{max} \) in the interference pattern occurs when the waves from the two slits interfere constructively, and is given by: \[ I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 \] Substituting the values of \( I_1 \) and \( I_2 \): \[ I_{max} = (\sqrt{I_0} + \sqrt{2I_0})^2 = (\sqrt{I_0} (1 + \sqrt{2}))^2 = I_0 (1 + \sqrt{2})^2 = I_0 (1 + 2 + 2\sqrt{2}) = I_0 (3 + 2\sqrt{2}) \] The minimum intensity \( I_{min} \) in the interference pattern occurs when the waves from the two slits interfere destructively, and is given by: \[ I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 \] Substituting the values of \( I_1 \) and \( I_2 \): \[ I_{min} = (\sqrt{I_0} - \sqrt{2I_0})^2 = (\sqrt{I_0} (1 - \sqrt{2}))^2 = I_0 (1 - \sqrt{2})^2 = I_0 (1 + 2 - 2\sqrt{2}) = I_0 (3 - 2\sqrt{2}) \] The ratio of the maximum to the minimum intensity is: \[ \frac{I_{max}}{I_{min}} = \frac{I_0 (3 + 2\sqrt{2})}{I_0 (3 - 2\sqrt{2})} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}} \] So, the ratio \( I_{max} : I_{min} \) is \( (3 + 2\sqrt{2}) : (3 - 2\sqrt{2}) \).

Answer 2

We need the ratio of maximum to minimum intensity in Young’s double-slit interference when one slit has half the width of the other. Let the wider slit have width \(w\) and the narrower slit have width \(w/2\).

Concept Used:

For two coherent sources producing intensities \(I_1\) and \(I_2\) (when each slit is opened alone), the resultant intensity on the screen is

\[ I=I_1+I_2+2\sqrt{I_1I_2}\cos\phi. \]

Hence, the extreme values are

\[ I_{\max}=(\sqrt{I_1}+\sqrt{I_2})^2,\qquad I_{\min}=(\sqrt{I_1}-\sqrt{I_2})^2. \]

With identical illumination and observation near the central region (same diffraction envelope), the intensity contributed by a single long slit is proportional to its width, i.e., \(I\propto w\).

Step-by-Step Solution:

Step 1: Assign single-slit intensities using proportionality to slit width.

\[ I_1 : I_2 \;=\; w : \frac{w}{2} \;=\; 2:1. \]

Step 2: Take square roots to obtain amplitude-like terms for the interference formula.

\[ \sqrt{I_1} : \sqrt{I_2} \;=\; \sqrt{2} : 1. \]

Step 3: Compute the extreme intensities using the standard relations.

\[ I_{\max}=(\sqrt{2}+1)^2 = 3+2\sqrt{2}, \] \[ I_{\min}=(\sqrt{2}-1)^2 = 3-2\sqrt{2}. \]

Final Computation & Result

Therefore, the ratio of maximum to minimum intensity is

\[ I_{\max} : I_{\min} \;=\; (3+2\sqrt{2}) : (3-2\sqrt{2}). \]

Correct Option: \((3 + 2\sqrt{2}) : (3 - 2\sqrt{2})\) (Option 2).