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In Young'S Double Slit Experiment, the Two Slits Are Separated by a Distance of 1.5 Mm and the Screen is Placed 1 M Away from the Plane of the Slits. - Physics

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प्रश्न

In Young's double-slit experiment, the two slits are separated by a distance of 1.5 mm, and the screen is placed 1 m away from the plane of the slits. A beam of light consisting of two wavelengths of 650 nm and 520 nm is used to obtain interference fringes.
Find the least distance from the central maximum where the bright fringes due to both the wavelengths coincide.

टिप्पणी लिखिए

उत्तर

Let n1 bright band of λ1 = 520 nm coincides with n2 bright band of λ2 = 650 nm

So `("n"_1lamda_1"D")/"d" = ("n"_2lamda_2"D")/"d"`

`"n"_1lamda_1 = "n"_2lamda_2`

`"n"_1/"n"_2 = lamda_2/lamda_1 = (50)/(520) = (5)/(4)`

so the least distance from the central maximum where the bright fringes due to both the wavelengths coincide is

`"x" = ("n"_1lamda_1"D")/"d" = (5 xx 520 xx 10^-9 xx 1)/(1.5 xx 10^-3)`

= `1.73 xx 10^-3 "m"`
= 1.73 mm

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Width of Central Maximum
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2014-2015 (March) Ajmer Set 2
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