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प्रश्न
In a biprism experiment, the fringes are observed in the focal plane of the eyepiece at a distance of 1.2 m from the slits. The distance between the central bright and the 20th bright band is 0.4 cm. When a convex lens is placed between the biprism and the eyepiece, 90 cm from the eyepiece, the distance between the two virtual magnified images is found to be 0.9 cm. Determine the wavelength of light used.
उत्तर १
Given:
D = 1.2 m
The distance between the central bright band and the 20th bright band is 0.4 cm.
∴ y20 = 0.4 cm = 0.4 × 10-2 m
W = `"y"_20/20 = 0.4/20 xx 10^-2 "m" = 2 xx 10^-4 "m"`,
d1 = 0.9 cm = 0.9 × 10-2 m, v1 = 90 cm = 0.9 m
∴ u1 = D - v1 = 1.2 m - 0.9 m = 0.3 m
Now, `"d"_1/"d" = "v"_1/"u"_1`
∴ d = `("d"_1"u"_1)/"v"_1 = ((0.9 xx 10^-2)(0.3))/0.9` m
= 3 × 10-3 m
∴ The wavelength of light,
`lambda = ("Wd"/"D") = (2 xx 10^-4 xx 3 xx 10^-3)/1.2` m
= 5 × 10-7 m
= 5 × 10-7 × 1010 Å
= 5000 Å
उत्तर २
Given:
- D = 1.2 m, X20 = 0.4 cm = 0.4 × 10-2 m,
- Image distance, ν = 90 cm = 90 × 10-2 m
- Object distance, u = 120 - 90 = 30 cm = 30 × 10-2 m
- Distance between two virtual images, d1 = 0.9 cm = 0.9 × 10-2 m
To find: λ = ?
We know that,
`"size of image"/"size of object" = "image distance"/"object distance"`
∴ `d_1/d = ν/u`
∴ `d = (d_1 xx u)/v = (0.9 xx 10^-2 xx 30 xx 10^-2)/(90 xx 10^-2) = 0.3 xx 10^-2` m
But, `X_n = (nlambdaD)/d`
∴ `0.4 xx 10^-2 = (20 xx lambda xx 1.2)/(0.3 xx 10^-2)`
∴ `lambda = (0.4 xx 10^-2 xx 0.3 xx 10^-2)/(20 xx 1.2)`
∴ `lambda = 5 xx 10^-7` m = 5000 Å
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