QJEEM23Ph350d: JEE Main 2023 Physics - Class 12 JEEM23Ph350d Solution: A single slit of width is illuminated by a monochromatic light…

When light passes through a narrow slit, it diffracts and produces a diffraction pattern on a screen. The pattern consists of a central bright maximum flanked by a series of alternating bright and dark fringes.

The condition for the first minimum in the diffraction pattern is given by :

$$a\sin \theta =\lambda$$

where $a$ is the width of the slit, $\theta$ is the angle of diffraction, and $\lambda$ is the wavelength of the incident light.

In this problem, we are given that the wavelength of the incident light is $\lambda = 600\,\mathrm{nm}$ , and the angle of diffraction for the first minimum is $\theta = 30^\circ$ . We want to find the width of the slit, $a$ , for which this occurs.

Substituting the given values into the condition for the first minimum, we get :

$$a\sin {30}^{\circ }=600\mathrm{nm}$$

Simplifying, we get :

$$a\cdot \frac{1}{2}=600\mathrm{nm}$$

Multiplying both sides by 2, we get :

$$a=1200\mathrm{nm}=\boxed{{\displaystyle 1.2\mu \mathrm{m}}}$$

Therefore, the width of the slit for which the first minimum appears at $\theta = 30^\circ$ is $\boxed{1.2\,\mu\mathrm{m}}$ .