Q43 / 107 · NTA Abhyas (JEE Main) 2020 · Physics · Wave Optics

First diffraction minimum due to a single slit of width $10^{- 4} c m$ is at $θ = 30^{o} .$ The wavelength of the light used is

A
$4000 Å$
Correct answer
B
$5000 Å$
C
$6000 Å$
D
$6250 Å$

Solution

a \sin θ = 1 \times λ

\Rightarrow λ = a \sin θ = 10^{- 4} \times \frac{1}{2} = \frac{10 \times 10^{- 5}}{2}

= 5 \times 10^{- 5} c m = 5000 Å

Hence, the correct answer is option: 'B'.

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