Question

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

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

Answer 1

The correct option is A $0.2mm$

Given
$\lambda =6\times {10}^{-7}m$
$D=50\times {10}^{-2}m$
$\Delta y=3\times {10}^{-3}m$

The position of $n^{th}$ minima in the diffraction pattern is
$\Rightarrow y_n=\frac{n\lambda D}{d}$

$\therefore$ Distance between $3^{rd}$ order minima and $1^{st}$ order minima will be,
$\Delta y=y_3-y_1=\frac{(3-1)\lambda D}{d}=\frac{2\lambda D}{d}=3mm$
Substituting known terms
$d=\frac{2(6\times {10}^{-7})(50\times {10}^{-2})}{3\times {10}^{-3}}$
$d=0.2mm$