Ch 34: Ray Optics

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 34: Ray Optics

Problem 79

Chapter 34, Problem 79

The mirror in FIGURE CP34.79 is covered with a piece of glass whose thickness at the center equals the mirror's radius of curvature. A point source of light is outside the glass. How far from the mirror is the image of this source?

Verified step by step guidance

Understand the problem: The mirror is covered with a piece of glass whose thickness at the center equals the mirror's radius of curvature. A point source of light is outside the glass, and we need to determine the location of the image formed by the mirror. This involves concepts of refraction, reflection, and optical path length.

Step 1: Identify the radius of curvature of the mirror (R). The radius of curvature is the distance from the mirror's surface to its center of curvature. This will be important for determining the focal length of the mirror, which is given by \( f = \frac{R}{2} \).

Step 2: Account for the glass covering the mirror. The glass introduces an optical path difference due to its refractive index \( n \). The effective optical path length through the glass is \( n \cdot t \), where \( t \) is the thickness of the glass. Since the thickness equals the radius of curvature \( R \), the optical path length becomes \( n \cdot R \).

Step 3: Use the mirror equation to find the image distance. The mirror equation is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o \) is the object distance (distance of the point source from the mirror), \( d_i \) is the image distance, and \( f \) is the focal length. Adjust \( f \) to account for the optical path difference introduced by the glass.

Step 4: Solve for \( d_i \) (image distance). Rearrange the mirror equation to isolate \( d_i \): \( d_i = \frac{1}{\frac{1}{f} - \frac{1}{d_o}} \). Substitute the adjusted focal length and the object distance into this equation to find the image distance.

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Key Concepts

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

Radius of Curvature

The radius of curvature is the distance from the mirror's surface to its center of curvature, which is the point where the mirror's surface would form a complete sphere. For spherical mirrors, this radius is crucial in determining how light rays reflect off the surface, influencing the focal point and image formation.

Refraction

Refraction is the bending of light as it passes from one medium to another, caused by a change in its speed. In this scenario, the glass covering the mirror introduces a new medium, affecting how light from the point source travels and ultimately impacts the perceived distance of the source from the mirror.

Image Formation in Mirrors

Image formation in mirrors involves the interaction of light rays with the mirror's surface, leading to the creation of virtual or real images. The position of the image depends on the object's distance from the mirror and the mirror's curvature, which is essential for determining how far the point source appears from the mirror.

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Mirror Equation

Related Practice

Textbook Question

CALC FIGURE CP34.81 shows a light ray that travels from point A to point B. The ray crosses the boundary at position x, making angles θ₁ and θ₂ in the two media. Suppose that you did not know Snell's law. You've proven that Snell's law is equivalent to the statement that 'light traveling between two points follows the path that requires the shortest time.' This interesting way of thinking about refraction is called Fermat's principle. Write an expression for the time t it takes the light ray to travel from A to B. Your expression should be in terms of the distances a, b, and w; the variable x; and the indices of refraction n₁ and n₂.

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Textbook Question

Some electro-optic materials can change their index of refraction in response to an applied voltage. Suppose a plano-convex lens (flat on one side, a 15.0 cm radius of curvature on the other), made from a material whose normal index of refraction is 1.500, is creating an image of an object that is 50.0 cm from the lens. By how much would the index of refraction need to be increased to move the image 5.0 cm closer to the lens?

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Textbook Question

A 2.0-cm-tall object is placed in front of a mirror. A 1.0-cm-tall upright image is formed behind the mirror, 150 cm from the object. What is the focal length of the mirror?

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Textbook Question

A fortune teller's 'crystal ball' (actually just glass) is 10 cm in diameter. Her secret ring is placed 6.0 cm from the edge of the ball. An image of the ring appears on the opposite side of the crystal ball. How far is the image from the center of the ball?

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Textbook Question

CALC A wildlife photographer with a 200-mm-focal-length telephoto lens on his camera is taking a picture of a rhinoceros that is 100 m away. Suddenly, the rhino starts charging straight toward the photographer at a speed of 5.0 m/s. What is the speed, in μm/s, image of the of the rhinoceros? Is the image moving toward or away from the lens?

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Textbook Question

Consider a lens having index of refraction n₂ and surfaces with radii R₁ and R₂. The lens is immersed in a fluid that has index of refraction n₁. A symmetric converging glass lens (i.e., two equally curved surfaces) has two surfaces with radii of 40 cm. Find the focal length of this lens in air and the focal length of this lens in water.

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