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 \(\mathrm{D}_{\text { }}\) is the distance between source and screen. The possible wavelengths used are proportional to

• A. \(\frac{1}{\mathrm{D}},\frac{1}{5\mathrm{D}},\frac{1}{7\mathrm{D}}\),
• B. \(\frac{1}{\mathrm{D}},\frac{1}{3\mathrm{D}},\frac{1}{5\mathrm{D}}\),
• C. \(\frac{1}{\mathrm{D}},\frac{1}{2\mathrm{D}},\frac{1}{3\mathrm{D}}\),
• D. \(\frac{1}{\mathrm{D}^{2}},\frac{1}{2\mathrm{D}^{2}},\frac{1}{3\mathrm{D}^{2}}\),

Hint:

Set the path difference at the point opposite one slit equal to an odd multiple of half a wavelength.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Point P is directly in front of slit \(S_1\). Path difference \(=S_2P-S_1P=\sqrt{D^{2}+d^{2}}-D\).

Step 2: Approximation
For \(D\gg d\): path difference \(\approx\dfrac{d^{2}}{2D}\).

Step 3: Minimum condition
\(\dfrac{d^{2}}{2D}=(2n-1)\dfrac{\lambda}{2}\Rightarrow\lambda=\dfrac{d^{2}}{(2n-1)D}\)

Step 4: Match
For \(n=1,2,3\): \(\lambda\propto\dfrac{1}{D},\dfrac{1}{3D},\dfrac{1}{5D}\), option (B).

Final Answer:
The possible wavelengths are proportional to 1/D, 1/3D, 1/5D. \[ \boxed{\text{(B) }\dfrac{1}{D},\dfrac{1}{3D},\dfrac{1}{5D}} \]