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Wave Optics

Question

In a Fraunhofer diffraction at single slit of width d with incident light of wavelength 5500 A^o, the first minimum is observed, at angle 30°. The first secondary maximum is observed at an angle θ =

sin^-1 $\frac{1}{\sqrt{2}}$

sin^-1 $\frac{1}{4}$

sin^-1 $\frac{3}{4}$

sin^-1 $\frac{\sqrt{3}}{2}$

Solution

sin^-1 $\frac{3}{4}$

Slit width = d

λ = 5500 A^o

A^o = 5.5 × 10^-7 m

θ_n = 30^o

For first secondary minima

d sin θ_n= λ

$d = \frac{λ}{\sin θ_{n}}$

⇒ d = $\frac{5.5 \times 10^{- 7}}{\sin 30^{o}}$

⇒ d = 11 × 10^-7m

For first secondary maxima

d sin θ_n= $\frac{3 λ}{2}$

⇒ sinθ_n = $\frac{3 λ}{2 d}$

= $\frac{3 \times 5.5 \times 10^{- 7}}{2 \times 11 \times 10^{- 7}}$

⇒ sinθ_n= $\frac{3}{4}$

⇒ θ_n= sin^-1 (3/4 )