Question

In Young's double slit experiment, two slits are illuminated with a light of wavelength \(λ\). The line joining \(A_1P\) is perpendicular to \(A_1A_2\) as shown in figure. If the first minimum is detected at P, the value of slits separation 'a' will be
(D = distance between source and screen)

• A. \(λ\text{D}\)
• B. \(\sqrt{λ\text{D}}\)
• C. \(\sqrt{\frac{λ}{\text{D}}}\)
• D. \(\sqrt{\frac{\text{D}}{λ}}\)

Hint:

The path difference at the first minimum is half a wavelength.

Answer 1

The Correct Option is B

Step 1: Geometry
From the figure, \(A_1P = D\) (perpendicular to the slit line) and \(A_2P = \sqrt{D^2+a^2}\).

Step 2: Condition for first minimum
Path difference \(= \frac\lambda2\): \(\sqrt{D^2+a^2} - D = \frac\lambda2\).

Step 3: Solve
\(D^2 + a^2 = D^2 + D\lambda + \frac{\lambda^2}{4}\), so \(a^2 = D\lambda + \frac{\lambda^2}4\). Since \(\lambda\ll D\), \(a^2\approx\lambda D\) and \(a = \sqrt{\lambda D}\). Option (B).

Final Answer:
The slit separation is root of lambda D. \[ \boxed{\text{(B)}\ a=\sqrt{\lambda D}} \]