Question

In Young's double slit experiment, in an interference pattern, a minimum is observed exactly in front of one slit. The distance between the two coherent sources is 'd' and 'D' is the distance between source and screen. The possible wavelengths used are inversely proportional to

• A. \(D,2D,3D,\ldots\)
• B. \(D,3D,5D,\ldots\)
• C. \(\frac{1}{D},\frac{2}{D},\frac{3}{D},\ldots\)
• D. \(\frac{1}{D^2},\frac{2}{D^2},\frac{3}{D^2},\ldots\)

Hint:

Set path difference d^2/(2D) equal to (2n - 1) lambda / 2.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A minimum occurs where the path difference is an odd multiple of half a wavelength: \(\Delta=(2n-1)\dfrac\lambda2\).

Step 2: Path difference at a point opposite one slit:
The point is at \(x=\dfrac d2\), so \(\Delta=\dfrac{xd}{D}=\dfrac{d^2}{2D}\).

Step 3: Solve for lambda:
\(\dfrac{d^2}{2D}=(2n-1)\dfrac\lambda2\), so \(\lambda=\dfrac{d^2}{(2n-1)D}\) with \(n=1,2,3,\ldots\)

Step 4: Read off the dependence:
\(\lambda\propto\dfrac{1}{(2n-1)D}\), so \(\lambda\) is inversely proportional to \(D,\ 3D,\ 5D,\ldots\) Option B.

Step 5: Why the other options are wrong.
Even multiples \(2D,4D\) would mean maxima, not minima. Options C and D give the wrong dependence on \(D\).

Final Answer:
Wavelength is inversely proportional to D, 3D, 5D, and so on. \[ \boxed{\text{(B) }D,\ 3D,\ 5D,\ldots} \]