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 38

Chapter 31, Problem 38

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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Identify the given values: The angle of incidence (θ₁) is 69°, and the refractive index of the glass (n₂) is 1.56. The refractive index of air (n₁) is approximately 1.00.

Recall Snell's Law, which relates the angles of incidence and refraction to the refractive indices:

n

₁⁢sin⁢θ1=n₂⁢sin⁢θ2, where θ₁ is the angle of incidence, θ₂ is the angle of refraction, n₁ is the refractive index of the first medium, and n₂ is the refractive index of the second medium.

Rearrange Snell's Law to solve for the sine of the angle of refraction (sin(θ₂)):

\sin θ_{2} = n

₁⁢sin⁢θ1n₂.

Substitute the known values into the equation:

\sin θ_{2} = \frac{1.00 \sin 69 °}{1.56}

. Calculate the sine of 69° and divide by 1.56.

Finally, use the inverse sine function (arcsin) to find the angle of refraction (θ₂):

θ_{2} = \arcsin \frac{1.00 \sin 69 °}{1.56}

. This will give you the angle of refraction in degrees.

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Key Concepts

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

Snell's Law

Snell's Law describes the relationship between the angles of incidence and refraction when light passes between two different media. It is mathematically expressed as n1 * sin(θ1) = n2 * sin(θ2), where n represents the refractive indices of the media and θ represents the angles relative to the normal. This law is fundamental for calculating how light bends when entering a new medium.

Refractive Index

The refractive index (n) of a material quantifies how much light slows down as it travels through that material compared to its speed in a vacuum. A higher refractive index indicates that light travels more slowly in that medium. In this question, the glass has a refractive index of 1.56, which is essential for applying Snell's Law to find the angle of refraction.

Angle of Incidence and Angle of Refraction

The angle of incidence is the angle between the incoming light ray and the normal line at the surface of the medium, while the angle of refraction is the angle between the refracted ray and the normal. These angles are crucial for applying Snell's Law, as they determine how much the light will bend when transitioning from one medium to another.

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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

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

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

When walking toward a concave mirror you notice that your image flips at a distance of 0.80 m from the mirror. What is the radius of curvature of the mirror? [Hint: Carefully examine Section 32–4.]

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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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