Solveeit Logo
Login

Question

Physics Question on Wave optics

In a Fraunhofer diffraction at single slit of width d with incident light of wavelength 5500 ??, the first minimum is observed, at angle 30∘^{\circ}. The first secondary maximum is observed at an angle θ\theta =

A

sin−112sin^{-1} \frac{1}{\sqrt{2}}

B

sin−114sin^{-1} \frac{1}{4}

C

sin−134sin^{-1} \frac{3}{4}

D

sin−132sin^{-1} \frac{\sqrt{3}}{2}

Answer

sin−134sin^{-1} \frac{3}{4}

Explanation

Solution

Slit width = d λ=5500?=5.5?10−7m,θn=30∘\lambda = 5500 ? = 5.5 ? 10^{-7} m, \theta_{n} = 30^{\circ} For first secondary minima, d sinθn=λ\theta_{n} = \lambda d=λsinθn=5.5×10−7sin30∘=11×10−7md=\frac{\lambda}{ sin\theta_{n}}=\frac{5.5\times10^{-7}}{sin 30^{\circ}}=11\times10^{-7}m For first secondary maxima, d sin θn=3λ2\theta_{n}=\frac{3\lambda}{2} i. e. sin θn=3λ2d=3×5.5×10−72×11×10−7\theta_{n}=\frac{3\lambda}{2d}=\frac{3\times5.5\times10^{-7}}{2\times11\times10^{-7}} sin θn=34\theta_{n}=\frac{3}{4} or θn=sin−1(3/4)\theta_{n}=sin^{-1} \left(3/4\right)