Question

In Young's double slit experiment, two slits \(A_1\) and \(A_2\) are illuminated with light of wavelength 600 nm. The line joining \(A_1O\) is perpendicular to \(A_1A_2\). If the first minimum is detected at 'O', the value of slit separation 'a' will be (D = 60 cm). [Figure: YDSE experimental setup with slits and screen]

• A. \(6\times10^{-4}\) m
• B. \(3\sqrt{10}\times10^{-2}\) cm
• C. \(\sqrt{10}\times10^{-6}\) cm
• D. \(6\times10^{-3}\) m

Hint:

The path difference at the point opposite one slit is \(\sqrt{D^2+a^2}-D\); set it equal to half a wavelength.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
O lies directly in front of \(A_1\), so \(A_1O=D\) and \(A_2O=\sqrt{D^2+a^2}\).

Step 2: First minimum condition
Path difference \(=\dfrac{\lambda}{2}\): \(\sqrt{D^2+a^2}-D=\dfrac{\lambda}{2}\).

Step 3: Simplify
Squaring gives \(a^2=D\lambda+\dfrac{\lambda^2}{4}\approx D\lambda\), since \(\lambda\ll D\).

Step 4: Calculation
\(a^2=0.6\times600\times10^{-9}=3.6\times10^{-7}\), so \(a=6\times10^{-4}\ \text{m}\).

Final Answer:
The slit separation is \(6\times10^{-4}\) m. \[ \boxed{\text{(A) }6\times10^{-4}\text{ m}} \]