Question

In Young's double slit experiment, in an interference pattern second minimum is observed exactly in front of one slit. The distance between the two coherent sources is ' \(d\) ' and the distance between source and screen is ' \(D\) '. The wavelength of light source used is

• A. \(\frac{d^{2}}{4D}\)
• B. \(\frac{d^{2}}{3D}\)
• C. \(\frac{d^{2}}{2D}\)
• D. \(\frac{d^{2}}{D}\)

Hint:

A point exactly in front of one slit is at distance d/2 from the central axis; use the path difference for the second minimum.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The point is in front of one slit, so its distance from the centre of the pattern is \(y=\dfrac{d}{2}\).

Step 2: Path difference
Path difference \(=\dfrac{yd}{D}=\dfrac{d}{2}\cdot\dfrac{d}{D}=\dfrac{d^{2}}{2D}\).

Step 3: Condition for second minimum
For the second dark fringe the path difference is \(\dfrac{3\lambda}{2}\).

Step 4: Calculation
\(\dfrac{3\lambda}{2}=\dfrac{d^{2}}{2D}\), so \(\lambda=\dfrac{d^{2}}{3D}\).

Final Answer:
The wavelength is \(\frac{d^{2}}{3D}\). \[ \boxed{\text{(B) }\dfrac{d^{2}}{3D}} \]