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प्रश्न
A beam of light consisting of two wavelengths 600 nm and 500 nm is used in Young's double slit experiment. The silt separation is 1.0 mm and the screen is kept 0.60 m away from the plane of the slits. Calculate:
- the distance of the second bright fringe from the central maximum for wavelength 500 nm, and
- the least distance from the central maximum where the bright fringes due to both wavelengths coincide.
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उत्तर
(i) Distance of 2nd bright fringe from the central maximum = `(2λ"D")/"d"`
= `(2 xx 500 xx 10^-9 xx 0.6)/(1 xx 10^-3)`
= 6 × 10−4 m
(ii) `("n"λ_1"D")/"d" = (("n" + 1)λ_2"D")/"d"`
Or, nλ1 = (n + 1)λ2
Or, `"n"/(("n" + 1)) = λ_2/λ_1`
Or, `"n"/(("n" + 1)) = 500/600`
∴ n = 5
So, least distance from central maximum = `(5 xx 600 xx 10^-9 xx 0.6)/(1 xx 10^-3)` = 18 × 10−4 m
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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.
REASON (R): Fringe width is proportional to (d/D).
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- Reason (R): The fringe width depends upon the wavelength of light (λ) used, the distance of the screen from the plane of slits (D) and slits separation (d).
