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Q.In a biprism experiment, light of wavelength 5200 Å is used to get an interference pattern on the screen. The fringe width changes by 1.3 mm when the screen is moved towards biprism by 50 cm. Find the distance between two virtual images of the slit.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2018Subjective· 3mImportance★★★★★
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Fringe width β=λD/d\beta = \lambda D/d; since λ\lambda and dd are fixed, a change in screen distance DD produces a proportional change in β\beta, letting us solve for dd.

In a biprism experiment, the fringe width is β=λDd\beta = \dfrac{\lambda D}{d}, where DD is the distance from the (virtual) slit images to the screen and dd is the separation between the two virtual images. Since λ\lambda and dd do not change as the screen is moved, the change in fringe width for a change in screen distance ΔD\Delta D is

Δβ=λ ΔDd  ⇒  d=λ ΔDΔβ.\Delta\beta = \frac{\lambda\,\Delta D}{d} \;\Rightarrow\; d = \frac{\lambda\,\Delta D}{\Delta\beta}. …

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