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Q.

Visible light of wavelength $6000 \times 10^{- 8}$ cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at $6 0^{0}$ from the central maximum. If the first minimum is produced at $θ_{1}$ , then $θ_{1}$ , is close to,

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a

$45^{0}$

b

$20^{0}$

c

$25^{0}$

d

$30^{0}$

answer is A.

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Detailed Solution

For 2nd minima

$d \sin θ = 2 λ$
$S i n c e, θ = 60^{0}, \sin θ = \frac{\sqrt{3}}{2} (g i v e n) \Rightarrow \frac{λ}{d} = \frac{\sqrt{3}}{4}______(i)$
So for 1st minima is
$d \sin θ_{1} = λ \Rightarrow \sin θ_{1} = \frac{λ}{d} = \frac{\sqrt{3}}{4} f r o m e q u a t i o n (1)$
$\Rightarrow θ = 25.65 º$
$\Rightarrow θ \approx 25 º$

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Visible light of wavelength 6000×10−8cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 600 from the central maximum. If the first minimum is produced at θ1, then θ1, is close to,