A single slit of width a is illuminated by a monochromatic light… — Solution | JEE Physics

MCQEasyJEE 2023Diffraction & Polarisation

A single slit of width $a$ is illuminated by a monochromatic light of wavelength $600\,nm$ . The value of $a$ for which the first minimum appears at $\theta = 30^\circ$ on the screen will be

A
$0.6\,\mum$
B
$1.2\,\mum$
C
$1.8\,\mum$
D
$3\,\mum$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Wavelength is $\lambda = 600\,nm = 0.6\,\mum$ and the first minimum appears at $\theta = 30^\circ$ .

Find: The slit width $a$ .

For single slit diffraction, the condition for the first minimum is

a \sin\theta = \lambda

Substituting the given values,

a = \frac{\lambda}{\sin\theta}

Now, $\sin 30^\circ = \frac{1}{2}$ , so

a = \frac{0.6}{\frac{1}{2}} = 1.2\,\mum

Therefore, the width of the slit is $1.2\,\mum$ and the correct option is B.

Common mistakes

Using the double slit interference formula instead of the single slit diffraction condition is incorrect. For the first minimum in single slit diffraction, use $a\sin\theta = \lambda$ , not an interference fringe relation.
Forgetting that $\sin 30^\circ = \frac{1}{2}$ leads to a wrong value of $a$ . Evaluate the trigonometric value first, then substitute carefully.
Not converting $600\,nm$ into $0.6\,\mum$ can cause unit inconsistency. Keep wavelength and slit width in the same unit before calculating.

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