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 85

Chapter 31, Problem 85

Light from a laser (in air) strikes the exact center of one face of a solid glass cube (n = 1.40) at an angle θ relative to the normal. The refracted beam travels inside the glass until it strikes an adjacent face of the cube. The original angle of incidence θ is such that no light exits the cube where the beam strikes the second face. What is the maximum value θ can have?

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Identify the relevant concepts: This problem involves Snell's Law, which relates the angles of incidence and refraction to the indices of refraction of the two media. The formula for Snell's Law is $ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) $, where $ n_1 $ and $ n_2 $ are the indices of refraction of the first and second media, respectively, and $ \theta_1 $ and $ \theta_2 $ are the angles of incidence and refraction.

Apply Snell's Law at the first interface (air to glass): Since the index of refraction of air is approximately 1 and that of glass is given as 1.40, use Snell's Law to find the angle of refraction $ \theta_2 $ inside the glass. The equation becomes $ \sin(\theta_2) = \frac{\sin(\theta)}{1.40} $.

Consider the internal reflection at the second face of the cube: For no light to exit the cube, the angle of incidence inside the cube must be greater than the critical angle for total internal reflection at the glass-air interface. The critical angle $ \theta_c $ can be calculated using $ \sin(\theta_c) = \frac{n_2}{n_1} $, where $ n_1 $ is the index of refraction of glass and $ n_2 $ is that of air.

Calculate the critical angle for total internal reflection: Substitute the indices of refraction into the formula for the critical angle to find $ \theta_c $.

Determine the maximum value of $ \theta $: The maximum value of $ \theta $ is such that the angle of incidence inside the glass ($ \theta_2 $) equals the critical angle. Use the relationship $ \sin(\theta_2) = \sin(\theta_c) $ and solve for $ \theta $ using Snell's Law.

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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 how light refracts when it passes from one medium to another, defined by the equation n1 * sin(θ1) = n2 * sin(θ2). Here, n represents the refractive index of the media, and θ represents the angle of incidence or refraction. This law is crucial for understanding how light behaves at the interface between air and glass, particularly in determining the angles involved in refraction.

Critical Angle

The critical angle is the angle of incidence above which total internal reflection occurs when light attempts to move from a denser medium to a less dense medium. It can be calculated using the formula θc = arcsin(n2/n1), where n1 is the refractive index of the denser medium and n2 is that of the less dense medium. In this scenario, the maximum angle θ must be less than the critical angle for light to remain within the glass cube.

Total Internal Reflection

Total internal reflection occurs when light traveling in a denser medium hits a boundary with a less dense medium at an angle greater than the critical angle, causing all the light to reflect back into the denser medium. This phenomenon is essential in understanding why the light beam does not exit the glass cube when it strikes the second face, as the angle of incidence must be carefully controlled to avoid exceeding the critical angle.

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Total Internal Reflection

Related Practice

Textbook Question

Suppose Fig. 32–37 shows a cylindrical rod whose end has a radius of curvature R = 2.0 cm, and the rod is immersed in water with index of refraction of 1.33. The rod has index of refraction 1.49. Find the location and height of the image of an object 2.0 mm high located 23 cm away from the rod.

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Textbook Question

A triangular prism made of crown glass (n = 1.52) with base angles of 26.0° is surrounded by air. If parallel rays are incident normally on its base as shown in Fig. 32–66, what is the angle Φ between the two emerging rays?

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Textbook Question

An object is placed 21 cm from a certain mirror. The image is half the height of the object, inverted, and real. How far is the image from the mirror, and what is the radius of curvature of the mirror?

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Textbook Question

The label on a laser says it produces light of wavelength 670 nm. The laser beam passes through a block of plastic for which n = 1.57. What is the wavelength of the light inside the plastic?

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Textbook Question

Figure 33–51 was taken from the NIST Laboratory (National Institute of Standards and Technology) in Boulder, CO, 2.0 km from the hiker in the photo. The Sun’s image was 15 mm across on the film. Estimate the focal length of the camera lens (actually a telescope). The Sun has diameter 1.4 x 10⁶ km, and it is 1.5 x 10⁸ km away.

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Textbook Question

When light passes through a prism, the angle that the refracted ray makes relative to the incident ray is called the deviation angle δ, Fig. 32–64. Show that this angle is a minimum when the ray passes through the prism symmetrically, perpendicular to the bisector of the apex angle Φ, and show that the minimum deviation angle, δ_m, is related to the prism’s index of refraction n by

$n = \frac{\sin \frac{1}{2} (ϕ + δ_{m})}{\sin ϕ / 2} .$

[Hint: For θ in radians, $(d / d θ) (\sin^{- 1} θ) = 1 / \sqrt{1 - θ^{2}}$ .]

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