(A) In single slit Fraunhofer diffraction,the angular width of the central maximum is given by $2\theta = \frac{2\lambda}{e}$,where $e$ is the slit wi

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(A) In single slit Fraunhofer diffraction,the angular width of the central maximum is given by $2\theta = \frac{2\lambda}{e}$,where $e$ is the slit width and $\lambda$ is the wavelength of light.From this relation,we can see that the angular half-width $\theta = \frac{\lambda}{e}$.Since $\theta$ is inversely proportional to the slit width $e$ (i.e.,$\theta \propto \frac{1}{e}$),decreasing the slit width $e$ will cause the value of $\theta$ to increase.Therefore,the correct option is $A$.

Class 12 Physics Wave Optics Single Slit Diffraction of Light

Easy

If we observe the single slit Fraunhofer diffraction with wavelength $\lambda$ and slit width $e$,the width of the central maxima is $2\theta$. On decreasing the slit width for the same $\lambda$,what happens to $\theta$?

A
$\theta$ increases
B
$\theta$ remains unchanged
C
$\theta$ decreases
D
$\theta$ increases or decreases depending on the intensity of light

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Class 12

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Wave Optics

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Single Slit Diffraction of Light

$A$ monochromatic light is incident on a single slit of width $0.014 \ mm$. The angular position of the second bright line observed is $2.81^{\circ}$. Then the wavelength of the incident light is $\left[\sin \left(2.81^{\circ}\right)=0.049072\right]$ (in $Å$)

MediumAP EAMCET 2022

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$A$ light wave of wavelength $\lambda$ is incident on a slit of width $d$. The resulting diffraction pattern is observed on a screen at a distance $D$. If the linear width of the principal maxima is equal to the width of the slit,then the distance $D$ is:

MediumMHT CET 2022

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$A$ screen is placed $50 \, cm$ from a single slit which is illuminated with $6000 \, \mathring{A}$. If the distance between the first and third minima in the diffraction pattern is $3 \, mm$,what is the width of the slit?

Medium

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$A$ light source with a wavelength of $5000 \, \mathring A$ produces a single-slit diffraction pattern. The first minimum in the diffraction pattern is observed at a distance of $5 \, mm$ from the central maximum. The distance between the slit and the screen is $2 \, m$. Find the width of the slit. (in $, mm$)

Easy

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Light of wavelength $5000 \, \mathring{A}$ is incident on a single slit such that the angular position of the first minima is $30^\circ$. The width of the slit is:

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If we observe the single slit Fraunhofer diffraction with wavelength $\lambda$ and slit width $e$,the width of the central maxima is $2\theta$. On decreasing the slit width for the same $\lambda$,what happens to $\theta$?

Similar Questions

$A$ monochromatic light is incident on a single slit of width $0.014 \ mm$. The angular position of the second bright line observed is $2.81^{\circ}$. Then the wavelength of the incident light is $\left[\sin \left(2.81^{\circ}\right)=0.049072\right]$ (in $Å$)

$A$ light wave of wavelength $\lambda$ is incident on a slit of width $d$. The resulting diffraction pattern is observed on a screen at a distance $D$. If the linear width of the principal maxima is equal to the width of the slit,then the distance $D$ is:

$A$ screen is placed $50 \, cm$ from a single slit which is illuminated with $6000 \, \mathring{A}$. If the distance between the first and third minima in the diffraction pattern is $3 \, mm$,what is the width of the slit?

Light of wavelength $5000 \, \mathring{A}$ is incident on a single slit such that the angular position of the first minima is $30^\circ$. The width of the slit is:

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