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:

The path difference at a point directly in front of a slit is d squared over 2D, and it must equal the condition for the required minimum.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For the n-th minimum the path difference is \(\frac{(2n-1)\lambda}{2}\). A point in front of one slit is at \(y=\frac{d}{2}\), where the path difference is \(\frac{yd}{D}=\frac{d^2}{2D}\).

Step 2: Third minimum
\(\frac{d^2}{2D}=\frac{5\lambda}{2}\).

Step 3: First minimum with new wavelength
\(\frac{d^2}{2D}=\frac{\lambda'}{2}\), so \(\lambda'=\frac{d^2}{D}=5\lambda\).

Step 4: Change in wavelength
The required change is \(\lambda'-\lambda=5\lambda-\lambda=4\lambda\).

Step 5: Compare the options
The change is \(4\lambda\), the third option.

Final Answer:
The wavelength must be increased by \(4\lambda\). \[ \boxed{\text{(C) }4\lambda} \]