Question

A screen is placed 50 cm from a single slit which is illuminated with light of wavelength 6000 $\stackrel{0}{A}$. If the distance between the first and third minima in the diffraction pattern is 3.0 mm, the width of the slit is

• A. $0.1mm$
• B. $0.2mm$
• C. $0.3mm$
• D. $0.4mm$

Answer 1

The correct option is B $0.2mm$

The position of $n^{th}$ minima in the diffraction pattern is
$x_n=\frac{nD\lambda }{d}$
$\therefore x_3-x_1=(3-1)\frac{D\lambda }{d}=\frac{2D\lambda }{d}$

$d=\frac{2D\lambda }{x_3-x_1}=\frac{2\times 0.50\times 6000\times {10}^{-10}}{3\times {10}^{-3}}=2\times {10}^{-4}m$