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Question

# A light ray, going through a prism with the angle of prism 60°, is found to deviate by 30°. What limit on the refractive index can be put from these data?

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Solution

## Given, The angle of the prism (A) = 60˚ The angle of deviation (δm) = 30˚ $\mathrm{Refractive}\mathrm{index},\phantom{\rule{0ex}{0ex}}\mathrm{\mu }\le \frac{\mathrm{sin}\left(\frac{A+{\delta }_{\mathrm{m}}}{2}\right)}{\mathrm{sin}\left(\frac{A}{2}\right)}\phantom{\rule{0ex}{0ex}}\mathrm{\mu }\le \frac{\mathrm{sin}\left(\frac{60°+{\delta }_{\mathrm{m}}}{2}\right)}{\mathrm{sin}\left(\frac{60°}{2}\right)}\phantom{\rule{0ex}{0ex}}\mathrm{\mu }\le 2\mathrm{sin}\left(\frac{60°+{\delta }_{\mathrm{m}}}{2}\right)$ As there is one ray that has been found which has deviated by 30˚, the angle of minimum deviation should be either equal to or less than 30˚ but it can not be more than 30˚. Therefore, $\mathrm{\mu }\le 2\mathrm{sin}\left(\frac{60°+{\delta }_{\mathrm{m}}}{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{\mu }\le 2\mathrm{sin}\left(\frac{60°+30°}{2}\right)=2\mathrm{sin}\left(45°\right)$ Refractive index (μ) will be more if angle of deviation (δm) is more. $\mathrm{\mu }\le 2×\frac{1}{\sqrt{2}}$ $⇒\mathrm{\mu }\le \sqrt{2}$ Hence, the required limit of refractive index is $\sqrt{2}$

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