(B) The condition for the first minimum in a single-slit Fraunhofer diffraction pattern is given by $a \sin \theta = n \lambda$. For the first minimum

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(B) The condition for the first minimum in a single-slit Fraunhofer diffraction pattern is given by $a \sin \theta = n \lambda$. For the first minimum,$n = 1$,so $\sin \theta = \frac{\lambda}{a}$.Here,the slit width $a = 12 \times 10^{-5} \text{ cm} = 12 \times 10^{-7} \text{ m}$.The wavelength $\lambda = 6000 \, \mathring{A} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m}$.Substituting the values: $\sin \theta = \frac{6 \times 10^{-7}}{12 \times 10^{-7}} = \frac{6}{12} = 0.5$.Since $\sin \theta = 0.5$,the half-angular width $\theta = \arcsin(0.5) = 30^{\circ}$.

Class 12 Physics Wave Optics Fresnel distance, Diffraction through circular slit, concept of Airy Disc, Rayleigh Criterion of Resolution , Fraunhofer diffraction

Medium

Find the half-angular width of the central maximum in the Fraunhofer diffraction pattern produced by a slit of width $12 \times 10^{-5} \text{ cm}$,when illuminated by monochromatic light of wavelength $6000 \, \mathring{A}$. (Answer in degrees)

A
$20$
B
$30$
C
$40$
D
$35$

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Wave Optics

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Fresnel distance, Diffraction through circular slit, concept of Airy Disc, Rayleigh Criterion of Resolution , Fraunhofer diffraction

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In Fraunhofer diffraction,light of wavelength $6328 \ \mathring{A}$ is incident perpendicularly on a slit of width $0.2 \ mm$. The angular width of the secondary maxima or minima will be: (in $^{\circ}$)

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Find the half-angular width of the central maximum in the Fraunhofer diffraction pattern produced by a slit of width $12 \times 10^{-5} \text{ cm}$,when illuminated by monochromatic light of wavelength $6000 \, \mathring{A}$. (Answer in degrees)

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