Question 92: Wave Optics

Question

A single slit of width $a$ is illuminated by a monochromatic light of wavelength $600 \mathrm{~nm}$. The value of ' $a$ ' for which first minimum appears at $	heta=30^{\circ}$ on the screen will be :

Accepted Answer

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 :
$$
a\sin	heta = \lambda
$$
where $a$ is the width of the slit, $	heta$ 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 $	heta = 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 :
$$
a = 1200\,\mathrm{nm} = \boxed{1.2\,\mu\mathrm{m}}
$$
Therefore, the width of the slit for which the first minimum appears at $	heta = 30^\circ$ is $\boxed{1.2\,\mu\mathrm{m}}$.

Answer

${3} \mu \mathrm{m}$

Answer

$0.6 \mu \mathrm{m}$

Answer

$1.8 \mu \mathrm{m}$

Answer

$1.2 \mu \mathrm{m}$

Wave Optics

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

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