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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
All textbooksGiancoli Douglas 5th editionCh. 32 - Light: Reflection and RefractionProblem 49
Chapter 31, Problem 49

(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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Start by understanding the geometry of the problem. A light ray enters a flat piece of glass at an angle θ (small angle approximation), refracts inside the glass, and exits on the other side. The goal is to calculate the perpendicular displacement d between the incident ray and the emerging ray.
Apply Snell's Law at the point of incidence: n1⁢sin⁢θ1=n2⁢sin⁢θ2. Here, n1 is the refractive index of air (approximately 1), and n2 is the refractive index of the glass (n). For small angles, use the approximation sin⁢θ≈θ, so θ2=θ1/n.
Inside the glass, the ray travels a distance t (thickness of the glass) at an angle θ2. The horizontal displacement of the ray inside the glass is given by x=t⁢tan⁢θ2. For small angles, use the approximation tan⁢θ≈θ, so x=t⁢θ2.
When the ray exits the glass, it refracts again. Using Snell's Law and the small angle approximation, the emerging ray is parallel to the incident ray but displaced by a perpendicular distance d. The displacement d is related to the horizontal displacement x by the factor d=x⁢(n-1)/n).
Substitute x=t⁢θ2 and θ2=θ/n into the expression for d. Simplify to get d=t⁢θ⁢(n-1)/n), which is the desired result.

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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 index of refraction. This phenomenon occurs due to the change in the speed of light in different materials. The relationship between the angles of incidence and refraction is described by Snell's Law, which states that n1 * sin(θ1) = n2 * sin(θ2), where n is the index of refraction and θ is the angle relative to the normal.
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Index of Refraction

The index of refraction (n) is a dimensionless number that describes how much light slows down in a medium compared to its speed in a vacuum. It is defined as n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the medium. A higher index indicates that light travels slower in that medium, affecting the angle at which it refracts when entering or exiting the material.
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Small Angle Approximation

The small angle approximation is a mathematical simplification used when angles are small (typically less than about 10 degrees). In this case, sin(θ) can be approximated as θ (in radians), which simplifies calculations involving trigonometric functions. This approximation is particularly useful in optics, where small angles are common, allowing for easier analysis of light behavior in systems like the one described in the question.
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Related Practice
Textbook Question

A parallel beam of light containing two wavelengths, λ₁ = 461 nm and λ₂ = 656 nm, enters the silicate flint glass of an equilateral prism as shown in Fig. 32–56. At what angle does each beam leave the prism (give angle with normal to the face)? See Fig. 32–28.

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Textbook Question

The critical angle for a certain liquid–air surface is 52.6°. What is the index of refraction of the liquid?

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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

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?

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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 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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