/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 A point object is placed at a di... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 57

A point object is placed at a distance of \(20 \mathrm{~cm}\) from a thin plano- convex lens of focal length \(15 \mathrm{~cm}\). If the plane surface is silvered, the image will form at (a) \(60 \mathrm{~cm}\) from left of \(A B\) (b) \(30 \mathrm{~cm}\) from left of \(A B\) (c) \(12 \mathrm{~cm}\) from left of \(A B\) (d) \(60 \mathrm{~cm}\) from right of \(A B\)

Short Answer

Expert verified
The image forms at approximately 12 cm from left of AB, thus (c).

Step by step solution

01

Understand Lens-Mirror System

In this problem, we are dealing with a plano-convex lens whose plane surface is silvered. This setup acts like a combination of a lens and a mirror, where the light first passes through the lens, reflects back from the silvered plane surface, and then passes through the lens again.
02

Use Lens Formula for First Passage

Use the lens formula \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \) for the first passage of light through the lens. Here, the object distance \( u = -20 \) cm (negative since object is on the opposite side of the incident light) and focal length \( f = 15 \) cm (positive for a converging lens). Substitute to find \( v \):\[ \frac{1}{15} = \frac{1}{v} - \frac{1}{-20} \]Solving gives:\[ \frac{1}{v} = \frac{1}{15} + \frac{1}{20} = \frac{4+3}{60} = \frac{7}{60} \] Thus, \( v = \frac{60}{7} \approx 8.57 \) cm.
03

Account for Reflection at Plane

The image formed at \( v = 8.57 \) cm by the lens acts as a virtual object for the mirror. The mirror's reflection will appear same distance away on the opposite side. Thus, the virtual object distance from the mirror is \( 8.57 \) cm to the left of the plane mirror.
04

Image Formation by Lens on Return Path

Now, consider this virtual object \( 8.57 \) cm left of the plane as a new object for the lens on the return path. Again use the lens formula with focal length \( f = 15 \) cm:\[ \frac{1}{15} = \frac{1}{v} - \frac{1}{-8.57} \]By solving it:\[ \frac{1}{v} = \frac{1}{15} + \frac{1}{8.57} \approx \frac{60}{857} + \frac{100}{857} = \frac{160}{857} \]Thus, \( v \approx 5.36 \) cm to the right of the plane mirror's initial position.
05

Find Image Distance from Reference Point AB

The reference point \( A B \) is initially at the plane mirror. So, the total distance of the final image from \( A B \) point is: initial object distance (20 cm) + image distance on return (5.36 cm). Thus, the image forms at approximately 25.36 cm left of \( A B \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Lens-Mirror System
In a lens-mirror system, a unique combination of optics is at play. We deal with a setup where a lens, specifically a plano-convex lens in this example, is paired with a mirror by silvering one side of the lens. This system acts like both a lens and a mirror, providing two opportunities for the light to be refracted and reflected.

Here's how it works:
  • First, light enters the lens, undergoing refraction and forming an initial image on the other side.
  • Then, the light reflects off the silvered plane (acting as a mirror).
  • Finally, the light passes through the lens once more, forming another image as it exits.
This makes a lens-mirror system particularly interesting, as it effectively duplicates the refractive and reflective processes.
Image Formation
Image formation within a lens-mirror system follows a sequence. Initially, light refracts through the lens to create an image at a specific location. This image then acts as a virtual object for the mirror aspect of the setup.

When studying image formation in a lens-mirror system:
  • The first image formed by the lens is treated as an object for the mirror.
  • The mirror reflects this image to form a new virtual object for the lens on its return journey.
  • Finally, the lens refracts this virtual object to produce the final image.
Understanding these steps helps to track where any image will eventually form, showcasing the power of combining lenses and mirrors. This precise sequence is vital for determining image location accurately.
Lens Formula
The lens formula is vital in determining the position of an image formed by a lens. In the context of our exercise, the lens formula is expressed as follows:\[\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\]

Where:
  • \( f \) is the focal length of the lens.
  • \( v \) is the image distance, or where the image forms.
  • \( u \) is the object distance, or where the initial object is placed.
For a converging lens, the focal length \( f \) is positive. When using this formula, keep in mind the sign conventions, such as taking distances in the direction of the incoming light as negative (for the initial object). Applying these conventions correctly is crucial to accurately finding where the image will form.
Focal Length
Focal length, represented by \( f \), is a fundamental property of a lens that determines its converging or diverging power. In a converging lens, like a plano-convex lens, the focal length is positive, indicating that it brings parallel rays of light to a focus.

This value is key in calculating:
  • The position of images via the lens formula.
  • The power of the lens, which is the inverse of the focal length, expressed in diopters \( (D = \frac{1}{f}) \).
A shorter focal length means a stronger lens, capable of bending light more significantly. In practical terms, understanding the focal length helps anticipate where and how sharply an image will be formed, a vital step in applications involving lens optics. Remember, in our particular exercise, the focal length of the lens being 15 cm is crucial in each stage of the calculation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In refraction, light waves are bent on passing from one medium to the second medium, because in the second medium (a) the frequency is different (b) the coefficient of elasticity is different (c) the speed is different (d) the amplitude is smaller

A concave mirror of focal length \(f\) produces an image \(n\) times the size of the object. If the image is real, then the distance of the object from the mirror is (a) \((n-1) f\) (b) \([(n-1) / n] f\) (c) \([(n+1) f / n]\) (d) \((n+1) f\)

If an equi-convex lens of focal length \(30 \mathrm{~cm}\) is cut into two equal parts by a horizontal plane, then (a) the light transmitting area of each part becomes half of the initial (b) the intensity will reduce to half (c) the aperture becomes \(\frac{1}{\sqrt{2}}\) times of its initial value (d) all of the above

The limit of resolution of an optical instrument arises on account of (a) reflection (b) diffraction (c) polarization (d) interference

A man suffering from short sight is unable to see objects distinctly at a distance greater than \(2 \mathrm{~m}\). The power of lens required to correct this defect should be (a) \(-0.5 \mathrm{D}\) (b) \(-2 \mathrm{D}\) (c) \(+0.5 \mathrm{D}\) \((\mathrm{d})+2 \mathrm{D}\)
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Fluids

Read Explanation

Magnetism

Read Explanation

Radiation

Read Explanation

Measurements

Read Explanation

Atoms and Radioactivity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.