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Question 75 of 83

Q.In a biprism experiment, the fringes are observed in the focal plane of the eye-piece at a distance of 1.2 m from the slit. The distance between the central bright band and the 20th bright band is 0.4 cm. When a convex lens is placed between the biprism and the eye-piece, 90 cm from the eye-piece, the distance between the two virtual magnified images is found to be 0.9 cm. Determine the wavelength of light used.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2023Subjective· 3mImportance★★★★★
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Combines the biprism fringe-width method with the lens-displacement method for source separation.

Fringe width from the un-lensed setup: with D=1.2D=1.2 m (slit to eyepiece) and 20 fringes spanning 0.4 cm from the centre,

β=0.420=0.02 cm\beta = \frac{0.4}{20} = 0.02\ \text{cm}

Source separation from the lens method: the lens is placed 90 cm from the eyepiece, so its distance from the biprism (source) is u=D−v=120−90=30u = D - v = 120-90 = 30 cm, with v=90v=90 cm. The magnification is

m=vu=9030=3m = \frac{v}{u} = \frac{90}{30} = 3

The two virtual images (magnified) are separated by d′=0.9d' = 0.9 cm, so the actual (unmagnified) separation of the two coherent virtual sources is …

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