First slide

Diffraction of light-Wave Optics

Question

Light of wavelength 6000 $\overset{0}{A}$ is incident normally on a slit of width $24 \times 10^{- 5}$ cm. Find out the angular position of second minimum from central maximum

Moderate

A

$53^{0}$
B

$37^{0}$
C

$30^{0}$
D

$60^{0}$

Solution

$F o r 2 n d m i n i m u m, a \sin θ = 2 λ p u t λ = 6 \times 10^{- 7} m, a = 24 \times 10^{- 5} \times 10^{- 2} m \Rightarrow \sin θ = \frac{2 λ}{a} = \frac{2 \times 6 \times 10^{- 7}}{24 \times 10^{- 7}} = \frac{1}{2} ∴ θ = 30^{0}$

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