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InterviewSolution
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The refractive indices of silicate fint glass for wavelength 400 nm and 700 nm are 1.66 aned 1.61 respectively. Find the minimum angles of deviation of an equailateral prism made of this glass for light of wavelength 400 nm and 700 nm. |
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Answer» The minumum angle of devision `delta_m` is given by `mu = (sin (A+delta_m)/(2))/(sin(A)/(2)) = (sin(30^@ + (delta_m)/(2)))/(sin [email protected])` `=2 sin(30^@ + (delta_m)/(2))` for 400 nm light, `1.66 = 2 sin (30^@ + delta_m//2)` or `sin(30^@ +delta_m//2) = 0.83` `or (30^@ + delta_m//2) = 56^@` For 700 nm light, `1.61 = 2 sin (30^@ + delta_m//2).` This gives `delta_m = 48^@` |
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