In a Young's double slits experiment, the ratio of amplitude of l… — Solution | JEE Physics

MCQEasyJEE 2023Interference (Young's Experiment)

In a Young's double slits experiment, the ratio of amplitude of light coming from slits is $2 : 1$ . The ratio of the maximum to minimum intensity in the interference pattern is:

A
$9 : 1$
B
$9 : 4$
C
$2 : 1$
D
$25 : 9$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The ratio of amplitudes is $\frac{A_1}{A_2} = \frac{2}{1}$ .

Find: The ratio $\frac{I_{\text{max}}}{I_{\text{min}}}$ in the interference pattern.

For two coherent sources,

\frac{I_{\text{max}}}{I_{\text{min}}} = \left( \frac{A_1 + A_2}{A_1 - A_2} \right)^2

Substituting $A_1 = 2$ and $A_2 = 1$ ,

\frac{I_{\text{max}}}{I_{\text{min}}} = \left( \frac{2 + 1}{2 - 1} \right)^2 = \left( \frac{3}{1} \right)^2 = 9:1

Therefore, the ratio of maximum to minimum intensity is $9 : 1$ . The correct option is A.

The solution labels option C, but the working clearly gives $9 : 1$ , which matches option A.

Common mistakes

Using intensity ratio directly as $2:1$ . This is wrong because the given ratio is of amplitudes, not intensities. First use the amplitude-based formula for interference extrema.
Using $\left( \frac{A_1 + A_2}{A_1} \right)^2$ or another incomplete expression. This is wrong because minimum intensity depends on destructive interference, so the denominator must involve $A_1 - A_2$ .
Forgetting to square the amplitude ratio expression. This is wrong because intensity is proportional to the square of amplitude. After forming $\frac{A_1 + A_2}{A_1 - A_2}$ , square the whole quantity.

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Answer

Step-by-step solution

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