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Question
Suppose a monochromatic X-ray beam of wavelength 100 pm is sent through a Young's double slit and the interference pattern is observed on a photographic plate placed 40 cm away from the slit. What should be the separation between the slits so that the successive maxima on the screen are separated by a distance of 0.1 mm?
Solution
Given:
`lambda = 10 "pm" = 100 xx 10^-12 "m"`
`D = 40 "cm" = 40 xx 10^-2 "m"`
`β = 0.1 "mm" = 0.1 xx 10^-3 "m"`
`β = (lambdaD)/d`
`d = (lambdaD)/β`
= `(100 xx 10^-12 xx 40 xx 10^-2)/(10^-3 xx 0.1)`
= `4 xx 10^-7 "m"`
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