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: Use the dark fringe formula:
Dark fringes lie at $x=\dfrac{(2n-1)\lambda D}{2d}$.

Step 2: Set x = d/2:
$\dfrac{(2n-1)\lambda D}{2d}=\dfrac d2$, so $\lambda=\dfrac{d^2}{(2n-1)D}$.

Step 3: Match:
The denominator takes the values $D,3D,5D,\ldots$, option B.

Final Answer:
Setting x = d/2 in the dark fringe formula gives lambda = d^2/((2n - 1) D). \[ \boxed{\text{(B) }D,\ 3D,\ 5D,\ldots} \]