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Q. Refractive index of glass is $ 1.520 $ for red light and $ 1.525 $ for blue light. Let $ D_1 $ and $ D_2 $ be angles of minimum deviation for red and blue light respectively in a prism of this glass. Then,

AMUAMU 2015Ray Optics and Optical Instruments

Solution:

The refractive index of a prism is given by
$\mu = \frac{\sin \frac{1}{2}\left(A+D\right)}{\sin \left(\frac{A}{D}\right)}\,\, ...\left(i\right)$
where, $A =$ angle of prism and
$D =$ angle of minimum deviation
It follows from Eq. $(i)$ that greater the refractive index, the greater is the angle of minimum deviation. Since, the refractive index for the blue light greater than that of red light, the angle of minimum deviation ($D_2$) for blue light will be greater than that (i.e. angle $D_1$) for red light.