(B) In Fresnel diffraction, the source of light and the screen are at finite distances from the obstacle or aperture. No lenses are required to make t

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the number of half-period zones that superpose at the point

(B) In Fresnel diffraction, the source of light and the screen are at finite distances from the obstacle or aperture. No lenses are required to make the rays parallel. The resulting diffraction pattern at any point on the screen is determined by the superposition of wavelets originating from different half-period zones of the wavefront. The intensity at a specific point depends on the number of half-period zones that contribute to the superposition at that point, as these zones can interfere constructively or destructively.

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

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Fresnel diffraction is produced due to light rays falling on a small obstacle. The intensity of light at a point on a screen beyond an obstacle depends on

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A
the focal length of lens used for observation
B
the number of half-period zones that superpose at the point
C
the square of the sum of the number of half period zones
D
the thickness of the obstacle

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Class 12

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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 the Fraunhofer diffraction experiment, $L$ is the distance between the screen and the obstacle, $b$ is the size of the obstacle, and $\lambda$ is the wavelength of the incident light. The general condition for the applicability of Fraunhofer diffraction is:

MediumAP EAMCET 2008

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In a Fraunhofer diffraction pattern,the slit width is $0.5 \ mm$ and the screen is at $2 \ m$ away from the lens. If the wavelength of the light used is $5500 \ Å$,then the distance between the first minima on either side of the central maximum is (assume $\theta$ is small and measured in radians). (in $mm$)

EasyMHT CET 2024

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What will be the angle of diffraction for the first minimum due to Fraunhofer diffraction with a source of light of wavelength $550 \,nm$ and a slit of width $0.55 \,mm$ (in $,rad$)?

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The distance for which ray optics becomes a good approximation for an aperture of $0.3 \ cm$ and a light of wavelength $6000 \ Å$ is (in $m$)

MediumTS EAMCET 2025

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$A$ single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light does the third secondary maximum in the diffraction pattern coincide with the second secondary maximum in the pattern for red light of wavelength $6500 Å$ (in $Å$)?

DifficultKCET 2007

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Fresnel diffraction is produced due to light rays falling on a small obstacle. The intensity of light at a point on a screen beyond an obstacle depends on

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In the Fraunhofer diffraction experiment, $L$ is the distance between the screen and the obstacle, $b$ is the size of the obstacle, and $\lambda$ is the wavelength of the incident light. The general condition for the applicability of Fraunhofer diffraction is:

What will be the angle of diffraction for the first minimum due to Fraunhofer diffraction with a source of light of wavelength $550 \,nm$ and a slit of width $0.55 \,mm$ (in $,rad$)?

The distance for which ray optics becomes a good approximation for an aperture of $0.3 \ cm$ and a light of wavelength $6000 \ Å$ is (in $m$)

$A$ single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light does the third secondary maximum in the diffraction pattern coincide with the second secondary maximum in the pattern for red light of wavelength $6500 Å$ (in $Å$)?

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