A beam of polychromatic light passes through a thin prism of prism angle 6^ . The refractive index of the material of the prism varies with wavelength ( ) as n( ) = + ^2 , where = 3\ m ^ -1 and = 0.096\ m ^2. If _ is the wavelength at which the angle of minimum deviation D_m is smallest, then the correct value of D_m at _ is
B. 4.8^
For a thin prism, the angle of deviation is given by D = (n - 1)A. To find the smallest value of deviation D_m, we need to find the minimum value of the refractive index n( ). Given n( ) = + ^2 . Differentiating n( ) with respect to and equating to zero for minimum: dn d = - 2 ^3 = 0 ^3 = 2 Substituting the given values = 3\ m ^ -1 and = 0.096\ m ^2: ^3 = 2 0.096 3 = 0.064 = 0.4\ m Now, substituting = 0.4\ m back into the expression for n( ): n_ = 3(0.4) + 0.096 (0.4)^2 n_ = 1.2 + 0.096 0.16 = 1.2 + 0.6 = 1.8 The smallest angle of deviation is: D_m = (n_ - 1)A D_m = (1.8 - 1) 6^ = 0.8 6^ = 4.8^ Answer: 4.8^
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