Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 34: Q. 78 (page 994)

A spherical mirror of radius R has its center at C, as shown in FIGURE P34.78. A ray parallel to the axis reflects through F, the focal point. Prove that f = R/2 if∅<<1rad.

Short Answer

Expert verified

The statement is proved.

Step by step solution

01

Given information

We have given,

Radius of the spherical mirror=R

∅<<1rad.

A ray incident parallel to the principle axis and reflects through F.

we have to provef=R2.

02

Simplify

According to the law of reflection

incident angle = Reflected angle

∠i=∠r

As incident angle is parallel to the principle axis.

then using the figure, we can say that the alternative angles are equal.

∠i=∠∅=∠r

since two angle of the triangle △CFAare equal then we can say that it is a isosceles triangle.

CF=FA

and CM=MA=R2

Cos∅=R2×CFCF=Rsec∅2

since pole to C length is R.

R=f+R2sec∅f=R-R2sec∅

since,

∅<<1rad∅≃0sec∅=1

then we write,

f=R-R2f=R2

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Shows a light ray that travels from point A to point B. The ray crosses the boundary at position x, making angles θ1and θ2in the two media. Suppose that you did not know Snell’s law.

A. 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

B. The time depends on x. There’s one value of x for which the light travels from A to B in the shortest possible time. We’ll call it xmin. Write an expression (but don’t try to solve it!) from which xmincould be found.

C. Now, by using the geometry of the figure, derive Snell’s law from your answer to part b.

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.

It’s night time, and you’ve dropped your goggles into a 3.0-m-deep swimming pool. If you hold a laser pointer 1.0 m above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.0 m from the edge. How far are the goggles from the edge of the pool?

Suppose you have two pinhole cameras. The first has a small round hole in the front. The second is identical except it has a square hole of the same area as the round hole in the first camera. Would the pictures taken by these two cameras, under the same conditions, be different in any obvious way? Explain.

The glass core of an optical fiber has an index of refraction1.60. The index of refraction of the cladding is1.48. What is the maximum angle a light ray can make with the wall of the core if it is to remain inside the fiber?

A light beam can use reflections to form a closed, N-sided polygon inside a solid, transparent cylinder if N is sufficiently large. What is the minimum possible value of N for light inside a cylinder of (a) water, (b) polystyrene plastic, and (c) cubic zirconia? Assume the cylinder is surrounded by air.
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Energy Physics

Read Explanation

Scientific Method Physics

Read Explanation

Magnetism

Read Explanation

Solid State Physics

Read Explanation

Mechanics and Materials

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.

Sign-up for free