Question

In Young's double slit experiment carried out with
light of wavelength $(\lambda )=5000\mathring{A}$ , the distance
between the slits is 0.2 mm and the screen is at 200 cm
from the slits. The central maximum is at .v = 0. The
third maximum (taking the central maximum as
zeroth maximum) will be at x equal to

• A. 1.67 cm
• B. 1.5 cm
• C. 0.5 cm
• D. 5.0 cm

Answer 1

Answer: (B)

$x=(n)\lambda \frac{D}{d}=3\times 5000\times {10}^{-10}\times \frac{2}{0.2\times {10}^{-3}}$
$=1.5\times {10}^{-2}m=1.5cm$