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: Understanding the Question:
The slit widths are in the ratio 4:1. We must find the ratio \(I_{max}/I_{min}\).

Step 2: Key Formula or Approach:
The intensity from a slit is proportional to its width. So \(I_1 : I_2 = 4 : 1\). Amplitude goes as the square root of intensity, so \(a_1 : a_2 = 2 : 1\). \[ I_{max} \propto (a_1 + a_2)^2, \qquad I_{min} \propto (a_1 - a_2)^2 \]

Step 3: Calculate:
Take \(a_1 = 2\) and \(a_2 = 1\). \[ I_{max} \propto (2+1)^2 = 9, \qquad I_{min} \propto (2-1)^2 = 1 \] \[ \frac{I_{max}}{I_{min}} = \frac{9}{1} \]

Step 4: Checking Each Option:
25:9 comes from wrongly using the width ratio 4:1 as the amplitude ratio. 16:1 comes from squaring the width ratio. 16:3 has no basis. Option 4 (9:1) is correct.

Final Answer:
The ratio of maximum to minimum intensity is 9 : 1, so option 4 is correct. \[ \boxed{9 : 1} \]