Question

In Young's double slit experiment let 'd' be the distance between two slits and 'D' be the distance between the slits and the screen. Using a monochromatic source of wavelength ' $\lambda$ ', in an interference pattern, third minimum is observed exactly in front of one of the slits. If at the same point on the screen first minimum is to be obtained, the required change in the wavelength is [ d&D are not changed].

• A. $2\lambda$
• B. $3\lambda$
• C. $4\lambda$
• D. $5\lambda$

Hint:

For the same position, wavelength is inversely proportional to the order of minimum $(2n-1)$.

Answer 1

The Correct Option is D

Step 1: Concept
Position of $n^{th}$ minimum is $y_n = (2n - 1)\frac{\lambda D}{2d}$.

Step 2: Meaning
Point in front of one slit means $y = d/2$.
Case 1 ($n=3$): $\frac{d}{2} = (2 \times 3 - 1)\frac{\lambda D}{2d} = \frac{5\lambda D}{2d} \implies d^2 = 5\lambda D$.

Step 3: Analysis
Case 2 ($n=1$, new $\lambda'$): $\frac{d}{2} = (2 \times 1 - 1)\frac{\lambda' D}{2d} = \frac{\lambda' D}{2d} \implies d^2 = \lambda' D$.
From equations: $\lambda' D = 5\lambda D \implies \lambda' = 5\lambda$.

Step 4: Conclusion
Change required $\Delta \lambda = \lambda' - \lambda = 4\lambda$. Final Answer: (D)