Explanation According to the question, when there is no liquid covering the hypotenuse of the prism,a total internal reflection of the light beam happens. For the total internal reflection,the incidence angle should be greater than the critical angle. So, the maximum allowable range of refractive index can be related to the critical angle. Given: The angle of a prism is 45 ° and the prism is right-angled.The Refractive index of a prism is n . Formula used: Snell’s law n 1 sin θ 1 = n 2 sin θ 2 where n 1 and n 2 are the refractive index of two mediums and θ 1 and θ 2 are incident angle and refracted angle. Angles in an isosceles right-angled triangle are 45 ° , 45 ° and 90 ° . Calculation: For deviation of the light beam at the right angle, the beam should undergo total internal reflection at the hypotenuse of the prism. For this, the critical angle for glass- liquid interface must be less than the angle of incidence that is 45 ° as described in the above diagram. The incidence angle and reflected angle is 45 ° by the angle concurrency theorem and perpendicular bisector theorem. θ c < 45 ° ... ( 1 ) According to Snell’s law, n sin θ c = n L sin 90 ° sin θ c = n L / n

7th Edition

ISBN: 9780321625915

Author: Douglas C. Giancoli

Publisher: PEARSON

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Chapter 23, Problem 86GP

To determine

The allowable range of index of refraction of a prism ( n ).

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