Question 377: Waves Optics

Question

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 $$\mathop A\limits^o $$. The width of the silt is

Accepted Answer

$$\begin{aligned} & 	ext { Distance between the first minima }=\frac{2 \lambda \mathrm{D}}{\mathrm{a}}=5 \mathrm{~mm}=5 	imes 10^{-3} \mathrm{~m} \ & \lambda=6000 \mathop A\limits^o =6 	imes 10^{-7} \mathrm{~m}, \mathrm{D}=80 \mathrm{~cm}=0.8 \mathrm{~m} \ & 	herefore \frac{2 	imes 6 	imes 10^{-7} 	imes 0.8}{\mathrm{a}}=5 	imes 10^{-3} \ & 	herefore \mathrm{a}=\frac{2 	imes 6 	imes 10^{-7} 	imes 0.8}{5 	imes 10^{-3}}=0.192 	imes 10^{-3} \mathrm{~m} \ & =0.192 \mathrm{~mm} \ & \end{aligned}$$

Answer

0.096 mm

Answer

0.576 mm

Answer

0.192 mm

Answer

0.384 mm

Waves Optics

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