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Question: In a Fraunhofer diffraction at single slit of width d with incident light of wavelength 5500A°, the ...

In a Fraunhofer diffraction at single slit of width d with incident light of wavelength 5500A°, the first minimum is observed at angle 30°. The first secondary maximum is observed at an angle θ=
A. sin⁡−1[12] B. sin⁡−1[14] C. sin⁡−1[34] D. sin⁡−1[32] \begin{aligned} & \text{A}\text{. }{{\sin }^{-1}}\left[ \dfrac{1}{\sqrt{2}} \right] \\\ & \text{B}\text{. }{{\sin }^{-1}}\left[ \dfrac{1}{4} \right] \\\ & \text{C}\text{. }{{\sin }^{-1}}\left[ \dfrac{3}{4} \right] \\\ & \text{D}\text{. }{{\sin }^{-1}}\left[ \dfrac{\sqrt{3}}{2} \right] \\\ \end{aligned}

Explanation

Solution

Hint: When light undergoes diffraction, the diffraction pattern is produced which consists of continuous maxima and minima. Here first we have to find the slit width using the condition of secondary minima and then should find angle θ of secondary maximum.

Complete step by step answer:
The diffraction pattern due to a single slit consists of a central bright band having alternate dark and weak bright bands of decreasing intensity on both sides.
Path difference for nth secondary minimum is given by –
dsin⁡θn=nλd\sin {{\theta }_{n}}=n\lambda where n=1,2,3 ...
first secondary minima, we have,
dsin⁡θ1=λd\sin {{\theta }_{1}}=\lambda
They have given wavelength of light,λ=5500A∘\lambda =5500A{}^\circ
Angle for first minimum = θ=30∘\theta =30{}^\circ . So,
d=λsin⁡θd=\dfrac{\lambda }{\sin \theta }
On substituting we get
d=5500×10−10sin⁡30∘ sin⁡30∘=0.5 \begin{aligned} & d=\dfrac{5500\times {{10}^{-10}}}{\sin {{30}^{{}^\circ }}} \\\ & \sin {{30}^{{}^\circ }}=0.5 \\\ \end{aligned}
∴d=11×10−7m\therefore d=11\times {{10}^{-7}}m
So now condition for nth secondary maxima is:
dsin⁡θn=(2n+1)λ2d\sin {{\theta }_{n}}=\left( 2n+1 \right)\dfrac{\lambda }{2}
for the first secondary maxima
dsin⁡θ=(2+1)λ2d\sin \theta =\left( 2+1 \right)\dfrac{\lambda }{2}
dsin⁡θn=3λ2 ∴sin⁡θn=3λ2d \begin{aligned} & d\sin {{\theta }_{n}}=\dfrac{3\lambda }{2} \\\ & \therefore \sin {{\theta }_{n}}=\dfrac{3\lambda }{2d} \\\ \end{aligned}
On substituting values, we get,
∴sin⁡θn=3×55×10−72×11×10−7 sin⁡θn=34 ⇒θn=sin⁡−1[34] \begin{aligned} & \therefore \sin {{\theta }_{n}}=\dfrac{3\times 55\times {{10}^{-7}}}{2\times 11\times {{10}^{-7}}} \\\ & \sin {{\theta }_{n}}=\dfrac{3}{4} \\\ & \Rightarrow {{\theta }_{n}}={{\sin }^{-1}}\left[ \dfrac{3}{4} \right] \\\ \end{aligned}
Thus, the correct option is C.

Note: Students may make mistakes by taking λ\lambda in place of λ2\dfrac{\lambda }{2} for secondary maxima.
May get confused between diffraction and interference phenomenon- diffraction is supposed to be due to interference of secondary wavelets from the exposed portion of wave front from the slit. In diffraction bright bands are of decreasing intensity whereas in interference all bright fringes have the same intensity.