(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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The diagrams below show the intensity distribution in diffraction of light of two sources. In which of the following cases the sources are just resolved?

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Two towers on top of two hills are $40 \, km$ apart. The line joining them passes $50 \, m$ above a hill halfway between the towers. What is the longest wavelength of radio waves (in $cm$),which can be sent between the towers without appreciable diffraction effects?

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The Fraunhofer spectrum is a

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$A$ Fraunhofer diffraction pattern due to a narrow slit is obtained on a screen placed at a distance $D$ from the slit whose slit width is $a$. The distance of the first secondary maximum from the central maximum is

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In Fraunhofer diffraction pattern,slit width is $0.3 \ mm$ and screen is at $1.5 \ m$ away from the lens. If wavelength of light used is $4500 \ Å$,then the distance between the first minimum on either side of the central maximum is [ $\theta$ is small and measured in radian.] (in $mm$)

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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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The diagrams below show the intensity distribution in diffraction of light of two sources. In which of the following cases the sources are just resolved?

Two towers on top of two hills are $40 \, km$ apart. The line joining them passes $50 \, m$ above a hill halfway between the towers. What is the longest wavelength of radio waves (in $cm$),which can be sent between the towers without appreciable diffraction effects?

The Fraunhofer spectrum is a

$A$ Fraunhofer diffraction pattern due to a narrow slit is obtained on a screen placed at a distance $D$ from the slit whose slit width is $a$. The distance of the first secondary maximum from the central maximum is

In Fraunhofer diffraction pattern,slit width is $0.3 \ mm$ and screen is at $1.5 \ m$ away from the lens. If wavelength of light used is $4500 \ Å$,then the distance between the first minimum on either side of the central maximum is [ $\theta$ is small and measured in radian.] (in $mm$)

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