JEE MAIN - Physics (2024 - 30th January Morning Shift - No. 5)
The diffraction pattern of a light of wavelength $$400 \mathrm{~nm}$$ diffracting from a slit of width $$0.2 \mathrm{~mm}$$ is focused on the focal plane of a convex lens of focal length $$100 \mathrm{~cm}$$. The width of the $$1^{\text {st }}$$ secondary maxima will be :
2 mm
0.2 mm
0.02 mm
2 cm
Explanation
Width of $$1^{\text {st }}$$ secondary maxima $$=\frac{\lambda}{a} \cdot D$$
Here
$$\begin{aligned} & a=0.2 \times 10^{-3} \mathrm{~m} \\ & \lambda=400 \times 10^{-9} \mathrm{~m} \\ & D=100 \times 10^{-2} \end{aligned}$$
Width of $$1^{\text {st }}$$ secondary maxima
$$\begin{aligned} & =\frac{400 \times 10^{-9}}{0.2 \times 10^{-3}} \times 100 \times 10^{-2} \\ & =2 \mathrm{~mm} \end{aligned}$$
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