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प्रश्न
In a biprism experiment, the light of wavelength 5200Å 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.
उत्तर
`lambda = 5200Å = 5.2 xx 10^(-7)m`
`X_1 - X_2 = 1.3mm = 1.3 xx 10^(-3) m`
`D_1 - D_2 = 50cm = 0.5m`
`X_1 - X_2 = (lambdaD_1)/d - (lambdaD_2)/d`
`=lambda/d (D_1 - D_2)`
`:. d = (lambda(D_1 - D_2))/(X_1-X_2)`
=` (5.2 xx 10^(-7)xx0.5)/(1.3 xx 10^(-3))`
= `2xx 10^(-4)m`
= 0.2 mm
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