(C) For the first minimum in a single slit diffraction pattern,the condition is given by $d \sin \theta = n \lambda$,where $n = 1$ for the first minim

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(C) For the first minimum in a single slit diffraction pattern,the condition is given by $d \sin \theta = n \lambda$,where $n = 1$ for the first minimum.Given: Wavelength $\lambda = 5000 \; \mathring{A} = 5000 \times 10^{-10} \; m = 5 \times 10^{-7} \; m$.Slit width $d = 0.001 \; mm = 10^{-3} \; mm = 10^{-6} \; m$.Substituting the values into the formula: $\sin \theta = \frac{\lambda}{d} = \frac{5 \times 10^{-7}}{10^{-6}} = 0.5$.Therefore,$\theta = \sin^{-1}(0.5) = 30^o$.

Class 12 Physics Wave Optics Single Slit Diffraction of Light

Easy

$A$ parallel beam of monochromatic light of wavelength $5000 \; \mathring{A}$ is incident normally on a single narrow slit of width $0.001 \; mm$. The light is focused by a convex lens on a screen placed on the focal plane. The first minimum will be formed for the angle of diffraction equal to ........$^o$.

AIPMT 1993 All AIPMT

A
$0$
B
$15$
C
$30$
D
$60$

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Wave Optics

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Single Slit Diffraction of Light

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The half angular width of the central maxima in the Fraunhofer diffraction due to a slit of width $\frac{1200}{\sqrt{2}} \ \mu m$ is $45^o$. Then the wavelength of the light is ...... $\mu m$.

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$A$ narrow slit of width $2 \, mm$ is illuminated by monochromatic light of wavelength $500 \, nm$. The distance between the first minima on either side on a screen at a distance of $1 \, m$ is ........ $mm$.

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In a diffraction experiment from a single slit,the angular width of the central maximum does $NOT$ depend upon:

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$A$ single slit diffraction pattern is obtained using a beam of red light. If the red light is replaced by blue light, then:

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