Wave Optics

JEE Main level, single correct, medium

142 of 191

Question 142 of 191

Visible light of wavelength 6000×10−8cm6000\times {10}^{-8}cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60o{60}^{o} from the central maximum. If the first minimum is produced at θ1,{\theta }_{1}, then θ1{\theta }_{1} is close to

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Answer: (C) 25o{25}^{o}

For 2nd{2}^{nd} minima,
dsin⁡⁡θ=2λd\sin ⁡\theta =2\lambda
sin⁡⁡θ=32\sin ⁡\theta =\dfrac{\sqrt{3}}{2} , (given)
⇒λd=34\Rightarrow \dfrac{\lambda }{d}=\dfrac{\sqrt{3}}{4} ...(1)
So, for 1st{1}^{st} minima
dsinθ=λdsin\theta =\lambda
⇒sin⁡θ=λd=34\Rightarrow \sin \theta =\dfrac{\lambda }{d}=\dfrac{\sqrt{3}}{4}
⇒θ=25.65o\Rightarrow \theta ={25.65}^{o}
⇒θ≈25o\Rightarrow \theta \approx {25}^{o}

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