Question

The maximum intensity in Young's double-slit experiment is I₀. Distance between the slits is

d=5l, where l is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance D = 10d?

• A. $\frac{I_{0}}{2}$
• B. $\frac{3}{4}I_{0}$
• C. I₀
• D. $\frac{I_{0}}{4}$

Answer 1

Answer: (A)

$\frac{I_{0}}{2}$

Answer 2

Dx at P = $\frac{dx}{D}$ = $\frac{d^{2}}{2D}$ = $\frac{(5\lambda)^{2}}{2 \times 10 \times d}$

Dx = $\frac{(5\lambda)^{2}}{2 \times 10 \times 5\lambda}$ = $\frac{\lambda}{4}$

Df = $\frac{2\pi}{\lambda}$×Dx = $\frac{\pi}{2}$

I₀ = 4I ̃ I = $\frac{I_{0}}{4}$

I_net = I + I + 2 $\sqrt{I}$ $\sqrt{I}$cos $\frac{\pi}{2}$ = 2I = $\frac{I_{0}}{2}$