Wave Optics

JEE Main level, single correct, medium

154 of 191

Question 154 of 191

The first diffraction minimum due to the single slit diffraction is seen at θ=30∘\theta=30^{\circ} for a light of wavelength 5000A˚5000 Å falling perpendicularly on the slit. The width of the slit is

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Answer: (C) 10×10−5 cm10 \times 10^{-5} \mathrm{~cm}

For first minimum, dsin⁡θ=λd \sin \theta=\lambda
⇒d=λsin⁡θ=5000×10−8 cmsin⁡30∘=5000×10−8 cm1/2=10×10−5 cm\begin{aligned} \Rightarrow \quad d & =\frac{\lambda}{\sin \theta}=\frac{5000 \times 10^{-8} \mathrm{~cm}}{\sin 30^{\circ}} \\ & =\frac{5000 \times 10^{-8} \mathrm{~cm}}{1 / 2}=10 \times 10^{-5} \mathrm{~cm} \end{aligned}

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