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Chapter 1: Problem 43

The path of a light beam in air goes from an angle of incidence of \(35^{\circ}\) to an angle of refraction of \(22^{\circ}\) when it enters a rectangular block of plastic. What is the index of refraction of the plastic?

Short Answer

Expert verified
The index of refraction of the plastic block is approximately 1.53, which is calculated using Snell's Law: \(n_2 = \frac{\sin{35^\circ}}{\sin{22^\circ}}\).

Step by step solution

01

Identify the given information

We are given: - Angle of incidence in air (\(i_1\)) = \(35^\circ\) - Angle of refraction in the plastic block (\(i_2\)) = \(22^\circ\) - Index of refraction of air (\(n_1\)) = 1 (approximately) - Index of refraction of the plastic block (\(n_2\)) = ?
02

Use Snell's Law to relate the angles and indices of refraction

The formula for Snell's Law is: \[n_1 \times \sin{i_1} = n_2 \times \sin{i_2}\] Plugging in the known values: \[1 \times \sin{35^\circ} = n_2 \times \sin{22^\circ}\]
03

Solve for the index of refraction of the plastic block

To find the index of refraction of the plastic (\(n_2\)), divide both sides of the equation by \(\sin{22^\circ}\): \[n_2 = \frac{\sin{35^\circ}}{\sin{22^\circ}}\] Now, calculate the value for \(n_2\): \[n_2 \approx \frac{\sin{35^\circ}}{\sin{22^\circ}} \approx 1.53\]
04

Interpret the result

The index of refraction of the plastic block is approximately 1.53. This indicates the extent to which the light beam is bent, or refracted, as it passes from air to plastic. The higher the index of refraction, the more the light beam is bent.

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Most popular questions from this chapter

What angle would the axis of a polarizing filter need to make with the direction of polarized light of intensity \(1.00 \mathrm{kW} / \mathrm{m}^{2}\) to reduce the intensity to \(10.0 \mathrm{W} / \mathrm{m}^{2} ?\)

A light beam in air has an angle of incidence of \(35^{\circ}\) at the surface of a glass plate. What are the angles of reflection and refraction?

Unreasonable results Light traveling from water to a gemstone strikes the surface at an angle of \(80.0^{\circ}\) and has an angle of refraction of \(15.2^{\circ} .\) (a) What is the speed of light in the gemstone? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

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Diffusion by reflection from a rough surface is described in this chapter. Light can also be diffused by refraction. Describe how this occurs in a specific situation, such as light interacting with crushed ice.
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