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Chapter 24: Problem 4

\(\bullet\) If you run away from a plane mirror at \(2.40 \mathrm{m} / \mathrm{s},\) at wha speed does your image move away from you?

Short Answer

Expert verified
Your image moves away at 4.80 m/s.

Step by step solution

01

Understand the Concept

When you move away from a plane mirror, your image appears to move away as well. The movement of the image in a plane mirror is twice the speed of your motion because the image moves at the same speed in the opposite direction.
02

Calculate Image Speed

Since you move away from the mirror at a speed of 2.40 m/s, your image will also appear to move away from the mirror at 2.40 m/s in the opposite direction. Hence, the speed at which the image moves away from you is twice your speed, 2.40 m/s + 2.40 m/s = 4.80 m/s.
03

Formulate the Result

Combining these insights, we see that both you and your image move apart from their mutual starting point at a combined speed of 4.80 m/s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Plane Mirror
A plane mirror is a flat mirror that reflects light, forming images of objects placed in front of it. When light strikes a plane mirror, it bounces back, creating a mirror image that appears to be the same distance behind the mirror as the object is in front. Plane mirrors are unique because they do not magnify or distort images.
They have practical uses such as in bathrooms, dressing rooms, and optical instruments. These mirrors provide a clear, undistorted image, making them ideal for daily visualization and reflection tasks.
Image Speed
In a plane mirror, the speed at which an image appears to move is closely tied to the speed of the object itself. When you move towards or away from the mirror, your image mimics this motion, creating an illusion of movement. Image speed in a mirror is calculated by considering the speed at which you move.
If you're moving away from the mirror at a velocity of 2.40 m/s, your image appears to move at the same speed in the opposite direction. To find out how fast your image seems to move away from you, double the speed of your motion. Therefore, the image speed relative to you is 4.80 m/s in this scenario.
Relative Motion
Relative motion refers to the motion of an object as observed from a particular frame of reference. In the context of a mirror, when you observe your reflection, you are essentially seeing how your movement compares to a stationary background. This means that even if you aren't conscious of your speed, the mirror allows you to perceive it.
When you move at a certain speed towards or away from a mirror, your image reflects that speed back towards or away from you at an equal rate. Understanding relative motion helps in predicting how objects or images move in relation to each other, whether it's in a mirror or in everyday experiences.
Physics Problem Solving
Solving physics problems involving mirrors requires a combination of conceptual understanding and mathematical calculations. Begin by clearly understanding the concept, like how a plane mirror creates twice the apparent motion for an image.
Then, use formulas or logical reasoning to compute the desired quantities, ensuring units are consistent. In this problem, recognizing that the image speed is double your own motion is crucial.
  • Break down the problem into smaller parts.
  • List known quantities, like speeds.
  • Understand the relationship between those quantities.
Finally, formulate the result by combining all insights, ensuring the outcome is logical and aligned with basic principles.

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Most popular questions from this chapter

A converging lens forms an image of an \(8.00-\) mm-tall real object. The image is 12.0 \(\mathrm{cm}\) to the left of the lens, 3.40 \(\mathrm{cm}\) tall, and erect. (a) What is the focal length of the lens? (b) Where is the object located? (c) Draw a principal-ray dia- gram for this situation.

\(\cdot\) The front, convex, surface of a lens made for eyeglasses has a radius of curvature of \(11.8 \mathrm{cm},\) and the back, concave, surface has a radius of curvature of 6.80 \(\mathrm{cm} .\) The index of refraction of the plastic lens material is 1.67 . Calculate the local length of the lens.

(. A converging meniscus lens (see Fig. 24.31\()\) with a refrac- tive index of 1.52 has spherical surfaces whose radii are 7.00 \(\mathrm{cm}\) and 4.00 \(\mathrm{cm} .\) What is the position of the image if an object is placed 24.0 \(\mathrm{cm}\) to the left of the lens? What is the magnification?

A diverging lens with a focal length of \(-48.0 \mathrm{cm}\) forms a virtual image 8.00 \(\mathrm{mm}\) tall, 17.0 \(\mathrm{cm}\) to the right of the lens. (a) Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? (b) Draw a principal-ray diagram for this situation.

\(\cdot\) To a person swimming 0.80 \(\mathrm{m}\) beneath the surface of the water in a swimming pool, the diving board directly overhead appears to be a height of 5.20 \(\mathrm{m}\) above the swimmer. What is the actual height of the diving board above the surface of the water?
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