Ch. 32 - Light: Reflection and Refraction

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 32 - Light: Reflection and Refraction

Problem 42

Chapter 31, Problem 42

A light beam strikes a 2.5-cm-thick piece of plastic with a refractive index of 1.62 at a 45° angle. The plastic is on top of a 3.8-cm-thick piece of glass for which n = 1.47. What is the distance D in Fig. 32–51?
$Light beam refracts at a 45° angle through plastic and glass layers, illustrating Snell's Law and distance D.$

Verified step by step guidance

Determine the angle of refraction in the plastic using Snell's Law: $ n_1 \sin \theta_1 = n_2 \sin \theta_2 $, where $ n_1 = 1.00 $ (air), $ \theta_1 = 45^\circ $, and $ n_2 = 1.62 $. Solve for $ \theta_2 $, the angle of refraction in the plastic.

Calculate the horizontal displacement of the light beam as it travels through the plastic. Use trigonometry: $ x_1 = t_1 \tan \theta_2 $, where $ t_1 = 2.5 \, \text{cm} $ is the thickness of the plastic and $ \theta_2 $ is the angle of refraction in the plastic.

Determine the angle of refraction in the glass using Snell's Law again: $ n_2 \sin \theta_2 = n_3 \sin \theta_3 $, where $ n_2 = 1.62 $, $ \theta_2 $ is the angle of incidence at the plastic-glass interface, $ n_3 = 1.47 $, and solve for $ \theta_3 $, the angle of refraction in the glass.

Calculate the horizontal displacement of the light beam as it travels through the glass. Use trigonometry: $ x_2 = t_2 \tan \theta_3 $, where $ t_2 = 3.8 \, \text{cm} $ is the thickness of the glass and $ \theta_3 $ is the angle of refraction in the glass.

Add the horizontal displacements from the plastic and the glass to find the total distance $ D $: $ D = x_1 + x_2 $.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Refraction

Refraction is the bending of light as it passes from one medium to another with a different refractive index. This phenomenon occurs because light travels at different speeds in different materials. The degree of bending can be described by Snell's Law, which relates the angles of incidence and refraction to the refractive indices of the two media.

Refractive Index

The refractive index (n) of a material is a dimensionless number that describes how much light slows down in that medium compared to a vacuum. A higher refractive index indicates that light travels more slowly in that material. In this problem, the refractive indices of plastic (1.62) and glass (1.47) are crucial for calculating the angles of refraction and the overall path of the light beam.

Geometry of Light Paths

Understanding the geometry of light paths is essential for solving problems involving refraction. This includes analyzing the angles at which light enters and exits different media, as well as the thickness of each layer. In this scenario, the thickness of the plastic and glass layers, along with the angles of incidence and refraction, will help determine the total distance D that the light travels through the materials.

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Related Practice

Textbook Question

(III) A light ray is incident on a flat piece of glass with index of refraction n as in Fig. 32–24. Show that if the incident angle θ is small, the emerging ray is displaced a distance d = tθ(n - 1)/n , where t is the thickness of the glass, θ is in radians, and d is the perpendicular distance between the incident ray and the (dashed) line of the emerging ray (Fig. 32–24).

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Textbook Question

(II) In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, 1.3 m above the water level, onto the surface of the water at a point 2.8 m from his foot at the edge of the pool (Fig. 32–53). Where does the spot of light hit the bottom of the pool which is 2.1 m deep? Measure from the bottom of the wall beneath his foot.

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Textbook Question

An object is placed a distance r in front of a wall, where r exactly equals the radius of curvature of a certain concave mirror. At what distance from the wall should this mirror be placed so that a real image of the object is formed on the wall? What is the lateral magnification of the image?

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Textbook Question

A light beam strikes a piece of glass at a 55.00° incident angle. The beam contains two wavelengths, 450.0 nm and 700.0 nm, for which the index of refraction of the glass is 1.4831 and 1.4754, respectively. What is the angle between the two refracted beams?

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Textbook Question

A shaving or makeup mirror is designed to magnify your face by a factor of 1.8 (when compared to a flat mirror) when your face is placed 20.0 cm in front of it.

(a) What type of mirror is it?

(b) Describe the type of image that it makes of your face.

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Textbook Question

A flashlight beam strikes the surface of a pane of glass (n = 1.56) at a 69° angle to the normal. What is the angle of refraction?

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