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 :
Show Hint
- \( (2\sqrt{2} + 1) : (2\sqrt{2} - 1) \)
- \( (3 + 2\sqrt{2}) : (3 - 2\sqrt{2}) \)
- \( 9 : 1 \)
- \( 3 : 1 \)
The Correct Option is B
Approach Solution - 1
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 \( (3 + 2\sqrt{2}) : (3 - 2\sqrt{2}) \).
Approach Solution -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).
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