/*! 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 43 The relation between lateral mag... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The relation between lateral magnification \(m\), object distance \(u\), and focal length \(f\) of a spherical mirror is (A) \(m=\frac{f-u}{f}\) (B) \(m=\frac{f}{f+u}\) (C) \(m=\frac{f+u}{f}\) (D) \(m=\frac{f}{f-u}\)

Short Answer

Expert verified
The correct relation between lateral magnification (m), object distance (u), and focal length (f) of a spherical mirror is (A) \( m = \frac{f-u}{f} \).

Step by step solution

01

Write down the mirror formula and the lateral magnification definition.

We have the mirror formula for a spherical mirror: \( \frac{1}{u}+\frac{1}{v}=\frac{1}{f} \) And the definition of lateral magnification (m): \( m = - \frac{v}{u} \)
02

Isolate v in the mirror formula.

To find the relation between m, u, and f, we will first isolate v in the mirror formula to get v in terms of u and f. \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \) Now, to get v, we'll take the reciprocal of both sides: \( v = \frac{1}{\frac{1}{f} - \frac{1}{u}} \)
03

Use the lateral magnification definition to eliminate v.

We will use the definition of lateral magnification, \( m = - \frac{v}{u} \), to eliminate v and find the required relation. First, isolate v in the lateral magnification formula: \( v = -mu \) Now, substitute this expression for v into the expression for v obtained in step 2: \( -mu = \frac{1}{\frac{1}{f} - \frac{1}{u}} \)
04

Simplify the expression for m:

Now, our goal is to simplify this expression to find the correct relation for m. First, multiply both sides by -u: \( mu = -u(\frac{1}{f - \frac{u}{u}}) \) Simplify the terms in the brackets on the right-hand side: \( mu = -u(\frac{1}{f - 1}) \) Finally, divide both sides by -u and rearrange to get the expression for m: \( m = \frac{-u}{f-u} \) Comparing the obtained relation with the given options, the answer is: (A) \( m=\frac{f-u}{f} \)

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 91Ó°ÊÓ!

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

Match the statement given in Column-I with those given in Column-II,(A) In refraction 1\. Speed of wave does not change (B) In reflection 2\. Wavelength is decreased (C) In refraction of a ray 3\. Frequency moving from rarer to a \(\quad\) doesn't change denser medium. (D) In reflection of a ray 4\. Phase change of moving in rarer from a \(\quad \pi\) radians takes denser medium. \(\quad\) place

\(A B C\) is a right-angled prism kept in air. A ray (1) is incident on the face \(A B\) along the normal. Refractive index of the material of prism is the minimum value that will be required so that ray (1) undergoes total internal reflection at the face \(A C\). Another ray ( 2 ) is incident on the face \(A B\) such that it emerges from face \(A C\) along the normal to \(A C\). A third ray (3) falls on the face \(B C\) and emerges from face \(A C\) such that its angle of emergence is the same as that of incidence. Assuming light \((1),(2)\), and (3) have the same wavelength, then match the following. \(A\)(A) Refractive index of the material 1\. 120 of prism is (B) Angle of incidence in degree of \(2.90\) ray \((2)\) is (C) Deviation in degree suffered by 3\. 2 ray (2) is (D) Deviation in degree suffered by 4\. 60 ray (3) is 5. \(1.5\)

In a Young's double slit experiment, the fringe width is found to be \(0.4 \mathrm{~mm}\). If the whole apparatus is immersed in water of refractive index (4/3), without disturbing the geometrical arrangement, the new fringe width will be (A) \(0.30 \mathrm{~mm}\) (B) \(0.40 \mathrm{~mm}\) (C) \(0.53 \mathrm{~mm}\) (D) 450 microns

An object is placed at \(20 \mathrm{~cm}\) from a convex mirror of focal length \(10 \mathrm{~cm}\). The image formed by the mirror is (A) real and at \(20 \mathrm{~cm}\) from the mirror. (B) virtual and at \(20 \mathrm{~cm}\) from the mirror. (C) virtual and at \((20 / 3) \mathrm{cm}\) from the mirror. (D) real and at ( \(20 / 3) \mathrm{cm}\) from the mirror.

On a hot summer night, the refractive index of air is smallest near the ground and increases with height from the ground. When a light beam is directed horizontally, the Huygens' principle leads us to conclude that as it travels, the light beam [2015] (A) goes horizontally without any deflection (B) bends downwards (C) bends upwards (D) becomes narrower

See all solutions

Recommended explanations on Physics Textbooks

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.