(C) For the first minima in single slit diffraction,the condition is given by $a \sin \theta = n \lambda$.For the first minima,$n = 1$,so $a \sin \the

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(C) For the first minima in single slit diffraction,the condition is given by $a \sin \theta = n \lambda$.For the first minima,$n = 1$,so $a \sin \theta = \lambda$.Given:Slit width $a = 0.001 \text{ mm} = 1 \times 10^{-6} \text{ m}$.Wavelength $\lambda = 5000 \mathring{A} = 5000 \times 10^{-10} \text{ m} = 5 \times 10^{-7} \text{ m}$.Substituting the values:$\sin \theta = \frac{\lambda}{a} = \frac{5 \times 10^{-7}}{1 \times 10^{-6}} = 0.5$.Since $\sin \theta = 0.5$,the angle of diffraction is $\theta = 30^{\circ}$.

Class 12 Physics Wave Optics Single Slit Diffraction of Light

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$A$ parallel beam of monochromatic light of wavelength $5000 \mathring{A}$ is incident normally on a single narrow slit of width $0.001 \text{ mm}$. The light is focused by a convex lens on a screen placed on its focal plane. The first minima will be formed for the angle of diffraction of . . . . . . (degree).

JEE MAIN 2024 All JEE MAIN

A
$40$
B
$20$
C
$30$
D
$10$

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Wave Optics

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Single Slit Diffraction of Light

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$A$ single slit diffraction pattern is formed with light of wavelength $6384 Å$. The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of wavelength $\lambda_0$. The value of $\lambda_0$ is (in $Å$)

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$A$ slit of size $0.15 \,cm$ is placed at $2.1 \,m$ from a screen. On illuminating it by a light of wavelength $5 \times 10^{-5} \,cm$,the width of the central maxima will be:

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The radius $r$ of the $n^{th}$ half-period zone is proportional to:

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Red light of wavelength $6500 \, \mathring{A}$ is incident on a slit of width $0.50 \, \text{mm}$. Find the distance between the two first minima on both sides of the central maximum in the diffraction pattern. The distance between the screen and the slit is $1.8 \, \text{m}$.

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$A$ single slit diffraction pattern is formed with light of wavelength $6384 Å$. The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of wavelength $\lambda_0$. The value of $\lambda_0$ is (in $Å$)

$A$ slit of size $0.15 \,cm$ is placed at $2.1 \,m$ from a screen. On illuminating it by a light of wavelength $5 \times 10^{-5} \,cm$,the width of the central maxima will be:

If a classroom door is open just a small amount, we can hear sounds coming from the room but we can't see what is going on inside the room because

The radius $r$ of the $n^{th}$ half-period zone is proportional to:

Red light of wavelength $6500 \, \mathring{A}$ is incident on a slit of width $0.50 \, \text{mm}$. Find the distance between the two first minima on both sides of the central maximum in the diffraction pattern. The distance between the screen and the slit is $1.8 \, \text{m}$.

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