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प्रश्न
In Young's double slit experiment, the distance of the 4th bright fringe from the centre of the interference pattern is 1.5 mm. The distance between the slits and the screen is 1.5 m, and the wavelength of light used is 500 nm. Calculate the distance between the two slits.
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उत्तर
Given that D = 1.5 m, λ = 500 nm = 500 × 10-9 m, n = 4, d = ?
In, YDSE, the formula for bright fringes (constructive interference) is
`"Y"_"n" = ("n"lambda"D")/"d"`
`Delta"Y" = (4 xx 500 xx 10^-9 xx 1.5)/"d" = 1.5 xx 10^-3`
⇒ `"d" = (4 xx 50 xx 10^-9 xx 1.5)/(1.5 xx 10^-3)`
⇒ d = 200 × 10-6 m
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Consider the arrangement shown in the figure. The distance D is large compared to the separation d between the slits.
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- Suppose d has this value. Find the distance x at which the next bright fringe is formed.
- Find the fringe-width.
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ASSERTION (A): In an interference pattern observed in Young's double slit experiment, if the separation (d) between coherent sources as well as the distance (D) of the screen from the coherent sources both are reduced to 1/3rd, then new fringe width remains the same.
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