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A child stands between two mirrors with his arms out, perpendicular to the mirrors. One plane mirror is \(5 \mathrm{m}\) away from his left hand and another plane mirror is \(7 \mathrm{m}\) away from his right hand. How far apart are the two images produced by the mirrors if the child has an arm span of \(0.5 \mathrm{m}\) ? (A) \(2 m\) (B) \(12 \mathrm{m}\) (C) \(12.5 \mathrm{m}\) (D) \(24.5 \mathrm{m}\)

Short Answer

Expert verified
The distance between the two images is \(24.5 \text{ m}\) (Option D).

Step by step solution

01

Calculate the distance to the first image

The distance to the image in the first mirror is twice the distance of the mirror from the left hand. This is because the image formed by a plane mirror appears to be the same distance behind the mirror as the object is in front of it. Therefore, the distance to the first image is \[2 \times 5 \text{ m} = 10 \text{ m}.\]
02

Calculate the distance to the second image

Similarly, the distance to the image in the second mirror is twice the distance of the mirror from the right hand. Therefore, the distance to the second image is \[2 \times 7 \text{ m} = 14 \text{ m}.\]
03

Determine the total distance between the two images

The total distance between the two images is the sum of the distance to the first image, the distance to the second image, and the arm span of the child. Hence, \[10 \text{ m} + 14 \text{ m} + 0.5 \text{ m} = 24.5 \text{ m}.\]

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Key Concepts

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

Plane Mirrors
Plane mirrors are flat mirrors that reflect images in their normal proportions but reversed from left to right. They are commonly found in bathrooms, bedrooms, and other areas where people use mirrors for grooming.
When you look into a plane mirror, the image you see is a virtual image. This means it appears to be behind the mirror. The light rays don't actually come from behind the mirror but travel to your eyes along paths that project back behind the mirror.
Plane mirrors always produce images that are:
  • Same size as the object
  • Same distance behind the mirror as the object is in front
  • Upright
  • Laterally inverted (left-right reversed)
Understanding how plane mirrors work is essential for solving problems related to mirror image formation.
Image Distance Calculation
Calculating the distance to an image formed by a plane mirror is straightforward. The image appears to be as far behind the mirror as the object is in front of it. If an object is positioned 5 meters in front of a plane mirror, the image will appear 5 meters behind the mirror.
In the exercise, the child is standing between two mirrors. The critical steps involved in the solution are:
  • Measure the distance from the object (child’s hand) to the mirror.
  • Double that distance to find how far the image appears to be behind the mirror.
For example, if the left hand is 5 meters from one mirror, the image formed by that mirror will appear 10 meters behind it. Similarly, if the right hand is 7 meters from the other mirror, the image will appear 14 meters behind the other mirror.
Geometric Optics
Geometric optics is the study of light in cases where the wave nature of light is not significant. It involves the principles of reflection and refraction.
With mirrors, geometric optics involves understanding how light rays reflect off surfaces and how images are formed. For plane mirrors, the Law of Reflection states that the angle of incidence equals the angle of reflection.
In the context of the exercise:
  • The image distance equals twice the distance of the object from the mirror.
  • Images formed by plane mirrors are always virtual and erect.
  • The sum of the distances for multiple reflections can explain the complete scenario.
By applying these principles, you can solve even complex problems involving multiple mirrors.
Mirror Image Properties
Understanding the properties of images formed by plane mirrors is important for solving problems. The properties include:
  • Virtual Images: The images cannot be projected on a screen because the light does not come from the image location.
  • Upright Images: The orientation of the image is the same as the object. If you stand upright, your image will also stand upright.
  • Same Size Images: The size of the image is the same as the object.
  • Laterally Inverted Images: The left side of the object appears as the right side in the image and vice versa.
In the exercise, knowing that the images are virtual, upright, and laterally inverted helps us combine the distances accurately. The total distance between the two images involves calculations that consider these properties.

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Most popular questions from this chapter

A submarine is inspecting the surface of the water with a laser that points from the submarine to the surface of the water and through the air. At what angle will the laser not penetrate the surface of the water but rather reflect entirely back into the water? (Assume \(n_{\text {water }}=1.33\) and \(n_{\text {air }}=1 .\) (A) \(19^{\circ}\) (B) \(29^{\circ}\) (C) \(39^{\circ}\) (D) \(49^{\circ}\)

An object is placed at the center of curvature of a concave mirror Which of the following is true about the image? (A) It is real and inverted. (B) It is virtual and inverted. (C) It is real and upright. (D) It is virtual and upright

A ray of light \(\left(f=5 \times 10^{14} \mathrm{Hz}\right)\) travels from air into crystal into chromium. If the indices of refraction of air, crystal, and chromium are \(1,2,\) and \(3,\) respectively, and the incident angle is \(30^{\circ},\) then which of the following describes the frequency and the angle of refraction in the chromium? (A) \(5 \times 10^{14} \mathrm{Hz} ; 9.6^{\circ}\) (B) \(5 \times 10^{14} \mathrm{Hz} ; 57^{\circ}\) (C) \(1.0 \times 10^{10} \mathrm{Hz} ; 9.6^{\circ}\) (D) \(1.0 \times 10^{10} \mathrm{Hz} ; 57^{\circ}\)

Which of the following describes the image formed by an object placed in front of a convex lens at a distance smaller than the focal length? (A) Virtual and inverted (B) Virtual and upright (C) Real and upright (D) Real and inverted

If a light ray has a frequency of \(5.0 \times 10^{14} \mathrm{Hz}\), in which region of the electromagnetic spectrum is it located? (A) X -ray (B) UV (C) Visible (D) Infrared

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