Question

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

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

Answer 1

The correct option is A 0.2 mm

Position of minima is given by,
$dsin\theta =n\lambda$
For small ${}^{\prime }{\theta }^{\prime },sin\theta \approx \theta =\frac{y}{D}$
$\Rightarrow y=\frac{n\lambda D}{d}$
$\therefore$ Distance between $3^{rd}$ order minima and $1^{st}$ order minima will be,
$\Delta y=y_3-y_1=\frac{(3-1)\lambda D}{d}=\frac{2\lambda D}{d}$
Substituting values, we get $\Delta y=0.2mm$