Question

In young double slit experiment $\frac{d}{D} = 10^{- 4}$ (d = distance between slits, D = distance of screen from the slits). At a point P on the screen resulting intensity is equal to the intensity due to individual slit I₀. Then the distance of point P from the central maxima is (λ = 6000 Å)

• A. 2 mm
• B. 1 mm
• C. 0.5 mm
• D. 4 mm

Answer 1

Answer: (A)

2 mm

Answer 2

By using shift $I = 4I_{0}\cos^{2}(\varphi/2)$ ⇒ $I_{0} = 4I_{0}\cos^{2}(\varphi/2)$ ⇒ $\cos(\varphi/2) = \frac{1}{2}$ or $\frac{\varphi}{2} = \frac{\pi}{3}$ ⇒ $\varphi = \frac{2\pi}{3}$

Also path difference $\Delta = \frac{xd}{D} = \frac{\lambda}{2\pi} \times \varphi$ ⇒

$x \times \left( \frac{d}{D} \right) = \frac{6000 \times 10^{- 10}}{2\pi} \times \frac{2\pi}{3}$ ⇒ $x = 2 \times 10^{- 3}m = 2mm.$