Question

If the two slits in a Young's double slit experiment have their widths in the ratio 4:1, then the ratio of intensities at maxima and minima in the interference pattern will be

• A. 16 : 3
• B. 16 : 1
• C. 25 : 9
• D. 9 : 1

Hint:

Intensity is proportional to width. Then \(I_{max}/I_{min} = (a_1+a_2)^2/(a_1-a_2)^2\).

Answer 1

The Correct Option is D

Step 1: Write with Intensity Directly:
Use $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$ and $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$. Widths are in ratio 4:1, so $I_1 = 4I_0$ and $I_2 = I_0$.

Step 2: Substitute:
$\sqrt{I_1} = 2\sqrt{I_0}$ and $\sqrt{I_2} = \sqrt{I_0}$. \[ I_{max} = (3\sqrt{I_0})^2 = 9I_0, \qquad I_{min} = (\sqrt{I_0})^2 = I_0 \]

Step 3: Take the Ratio:
\[ \frac{I_{max}}{I_{min}} = \frac{9I_0}{I_0} = 9 \] This is 9 : 1.

Step 4: Match:
9 : 1 is the fourth printed option.

Final Answer:
\[\boxed{9 : 1\ \text{(option 4)}}\]