Question

In Young’s double slit experiment, the 6th maximum with wavelength \( \lambda_1 \) is at a distance \( d_1 \) from the central maximum and the 4th maximum with wavelength \( \lambda_2 \) is at distance \( d_2 \). Then \( \dfrac{d_1}{d_2} \) is

• A. \( \dfrac{2\lambda_1}{3\lambda_2} \)
• B. \( \dfrac{3\lambda_1}{2\lambda_2} \)
• C. \( \dfrac{2\lambda_2}{3\lambda_1} \)
• D. \( \dfrac{3\lambda_2}{2\lambda_1} \)

Hint:

In YDSE, fringe position is directly proportional to the order of maximum and wavelength.

Answer 1

The Correct Option is B

Step 1: Position of nth maximum.
\[ y_n = \frac{n\lambda D}{d} \]

Step 2: Write expressions for given maxima.
\[ d_1 = \frac{6\lambda_1 D}{d}, \quad d_2 = \frac{4\lambda_2 D}{d} \]

Step 3: Take ratio.
\[ \frac{d_1}{d_2} = \frac{6\lambda_1}{4\lambda_2} = \frac{3\lambda_1}{2\lambda_2} \]