In a single slit diffraction pattern, the distance between the first minimum on the left and the first…
In a single slit diffraction pattern, the distance between the first minimum on the left and the first minimum on the right is $5 \mathrm{~mm}$. The screen on which the diffraction pattern is obtained is at a distance of $80 \mathrm{~cm}$ from the slit. The wavelength used is $6000 Å$. The width of the silt is
$0.096 \mathrm{~mm}$
$0.576 \mathrm{~mm}$
$0.192 \mathrm{~mm}$
$0.384 \mathrm{~mm}$
Solution
$\begin{aligned} & \text { Distance between the first minima on both side }=\frac{2 \lambda D}{a} \\ & =5 \times 10^{-3} \mathrm{~m} \frac{2 \times 6 \times 10^{-7} \times 0.8}{\mathrm{a}}=5 \times 10^{-3} \\ & \therefore \mathrm{a}=\frac{2 \times 6 \times 10^{-7} \times 0.8}{5 \times 10^{-3}}=0.192 \times 10^{-3} \mathrm{~m} \\ & =0.192 \mathrm{~mm}\end{aligned}$