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A monochromatic light of wavelength 500 ...

A monochromatic light of wavelength `500 nm` is incident normally on a single slit of width `0.2 mm` to produce a diffraction pattern. Find the angular width of central maximum obtained on the screen, `1 m` away.
Estimate the number of fringes obtained in YDSE with fringe width `0.5 mm`, which can be accommodated within the region of total angular spread of the central maximum due to a single slit.

Text Solution

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Calculation of angular width of central maxima Estimation of number of fringes.
Angular width of central maximum
`omega=(2lamda)/a`
`=(2xx5xx10^(-9))/(0.2xx10^(-3))` radin
`=5xx10^(-3)` radin
`beta=(lamdaD)/d`
Lines width of central maxima in the diffraction pattern 2XD
`omega'=(2lamdaD)/a`
Let n be the number of interference fringes which can be accommodated in the central maxima.
`:. nxxbeta= `
[Award the last 5 mark if the student writes the answer as 2 (taking `d=a`), or just attempts to do these calculation.]
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