/*! 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} Q42P A ray of light travelling in air... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A ray of light travelling in air is incident at angle uaon one face of a prism made of glass. Part of the light refracts into the prism and strikes the opposite face at point A(Fig. P33.42). If the ray at Ais at the critical angle, what is the value of?

Short Answer

Expert verified

the index of refraction is 1.305

the critical angle is 57.05°

Step by step solution

01

About Critical angle 

The critical angle is the angle of incidence where the angle of refraction is 90°. The light must travel from an optically more dense medium to an optically less dense medium.

02

Determine the index of refraction 

We have the following prism with incident ray at an angle of 0“ from the normal to the prism surface. Since the prism is

having an angle of 90° at point c In triangle ABC A CBA would be 50° and A CAB would be 40°- 80 We can see that ray will

have an incidence angle of 50 ° at face AC. question is saying that at point A light is getting total internal re?ection so angle

after refraction Would be 90° -

Assume the index of refraction of prism equal to n- Apply the Snell's Law at point A-

Therefore the index of refraction is 1.305

03

Determine the Critical angle 

Now we got the index of refraction of prism apply Snell's Law again at point B to get the value of 00-

Therefore the critical angle is 57.05°

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

Light requires about 8 minutes to travel from the sun to the earth. Is it delayed appreciably by the earth’s atmosphere? Explain.

Huygens’s principle also applies to sound waves. During the day, the temperature of the atmosphere decreases with increasing altitude above the ground. But at night, when the ground cools, there is a layer of air just above the surface in which the temperature increaseswith altitude. Use this to explain why sound waves from distant sources can be heard more clearly at night than in the daytime. (Hint:The speed of sound increases with increasing temperature. Use the ideas displayed in Fig. 33.36 for light.)

A student claimed that, because of atmospheric refraction (see Discussion Question Q33.2), the sun can be seen after it has set and that the day is therefore longer than it would be if the earth had no atmosphere. First, what does she mean by saying that the sun can be seen after it has set? Second, comment on the validity of her conclusion.

An interference pattern is produced by eight equally spaced narrow slits. The caption for Fig. 36.14 claims that minima occur for f =3π/4, π/4, 3π/2 and7π/4. Draw the phasor diagram for each of these four cases, and explain why each diagram proves that there is in fact a minimum. In each case, for which pairs of slits are there totally destructive interference?

A person looks at his reflection in the concave side of a shiny spoon. Is it right side up or inverted? Does it matter how far his face is from the spoon? What if he looks in the convex side?

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.