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Chapter 30: Problem 18

A light ray propagates in a transparent material at \(15^{\circ}\) to the normal to the surface. It emerges into the surrounding air at \(24^{\circ}\) to the normal. Find the material's refractive index.

Short Answer

Expert verified
The refractive index of the material is 0.76.

Step by step solution

01

Identify the known variables

The angle of incidence, \(i = 15^{\circ}\), and the angle of refraction, \(r = 24^{\circ}\). Also, we know the refractive index of air is 1.
02

Convert the angles into radians

Because sine functions in calculators use radians, convert both \(i\) and \(r\) into radians: \(i = 15^{\circ} * \pi/180 = 0.2618\) rad and \(r = 24^{\circ} * \pi/180 = 0.4189\) rad.
03

Apply Snell's law to find the refractive index

Snell's law states \(n_{1} * \sin{(i)} = n_{2} * \sin{(r)}\). We are looking for \(n_{2}\), the refractive index of the material, so we rearrange the equation to solve for \(n_{2}\): \(n_{2} = \frac{n_{1} * \sin{(i)}}{\sin{(r)}}\).
04

Calculate the refractive index

Considering that \(n_{1}\) for air is 1, we find: \(n_{2} = \frac{1 * \sin{(0.2618)}}{\sin{(0.4189)}} = 0.7598\). As refractive indices are often listed to two decimal places, we get \(n_{2} = 0.76\).

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Most popular questions from this chapter

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