Question

In Young's double slit experiment, the intensity of light coming from one of the slits is double the intensity from the other slit. The ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed is

• A. 34
• B. 40
• C. 25
• D. 38

Hint:

$I_{max}/I_{min} = (\sqrt{I_1/I_2} + 1)^2/(\sqrt{I_1/I_2} - 1)^2$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
$I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$, $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
Step 2: Detailed Explanation:
Let $I_1 = I_0$, $I_2 = 2I_0$. Then $I_{max} = (\sqrt{I_0} + \sqrt{2I_0})^2 = I_0(1 + \sqrt{2})^2 = I_0(3 + 2\sqrt{2})$. $I_{min} = (\sqrt{I_0} - \sqrt{2I_0})^2 = I_0(1 - \sqrt{2})^2 = I_0(3 - 2\sqrt{2})$. Ratio $\frac{I_{max}}{I_{min}} = \frac{3 + 2\sqrt{2}}{3 - 2\sqrt{2}} = (3 + 2\sqrt{2})^2 = 9 + 12\sqrt{2} + 8 = 17 + 12\sqrt{2} \approx 17 + 16.97 = 33.97 \approx 34$.
Step 3: Final Answer:
The ratio is 34.