Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A narrow beam of white light enters a prism made of crown glass at a 45.0° incident angle, as shown in Figure 25.57. At what angles, θRand θV, do the red (660nm) and violet (410nm) components of the light emerge from the prism?


Short Answer

Expert verified

The angle at which the red light emerges from the prism is obtained: QR=53.5°.

The angle at which the violet light emerges from the prism is obtained: QV=55.2.

Step by step solution

01

Define Geometric Optics

Geometrical optics, often known as ray optics, is an optics model that describes light propagation using rays. In geometric optics, a ray is an abstraction that can be used to approximate the routes along which light propagates under particular conditions.

02

Evaluating the angle of refraction

The light is incident at an angle of the value a. It undergoes refraction inside the prism. Then assuming the angle of refraction inside the prism be βand γ.

So, we obtain:

(90°β)+(90°γ)+60°=180°180°βγ+60°=180°γ=60°β

For the red light, we have: .n2=1.512

For the violet light, we have: n2=1.530.

03

Evaluating the angle made by the red light

Using the Snell’s law, we obtain:

n1sinα=n2sinββ=sin1n,sinαn2β=sin11×sin45°1.512β=sin10.4677β=27.883

With the help of the above equation, we have:

γ=60°-βγ=60-270883°γ=32.117°

04

Evaluating the angle of the red light making prism

Applying the Snell’s law and we obtain:

n2sinγ=n1sinθRθR=sin1n2sinγθR=sin11.512×sin32.117°θR=53.5°

Therefore, the angle at which red light emerges from the prism is:QR=53.5

05

Evaluating the angle made by violet light

Using the Snell’s law, we obtain:

n1sinα=n2sinββ=sin1n1sinαn2β=sin11×sin45°1.530β=27.527

With the help of the above equation, we have:

γ=60-βγ=60-27.527°γ=32.473

06

Evaluating the angle of the violet light making prism

Applying the Snell’s law and we obtain:

(n2)sinγ=n1sinθvθv=sin1(n2sinγ)θv=sin11.53×sin32.473θv=55.2

Therefore, the angle at which violet light emerges from the prism is: QV=55.2.

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 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

There was a major collision of an asteroid with the Moon in medieval times. It was described by monks at Canterbury Cathedral in England as a red glow on and around the Moon. How long after the asteroid hit the Moon, which is 3.84 x 105 kmaway, would the light first arrive on Earth?

It can be argued that a flat piece of glass, such as in a window, is like a lens with an infinite focal length. If so, where does it form an image? That is, how aredianddorelated?

Construct Your Own Problem Consider sunlight entering the Earth’s atmosphere at sunrise and sunset—that is, at a 90º incident angle. Taking the boundary between nearly empty space and the atmosphere to be sudden, calculate the angle of refraction for sunlight. This lengthens the time the Sun appears to be above the horizon, both at sunrise and sunset. Now construct a problem in which you determine the angle of refraction for different models of the atmosphere, such as various layers of varying density. Your instructor may wish to guide you on the level of complexity to consider and on how the index of refraction varies with air density.

A camera with a \(100{\rm{ }}mm\)focal length lens is used to photograph the sun and moon. What is the height of the image of the sun on the film, given the sun is \(1.40 \times {10^6}{\rm{ }}km\) in diameter and is \(1.50 \times {10^8}{\rm{ }}km\) away?

You can determine the index of refraction of a substance by determining its critical angle.

(a) What is the index of refraction of a substance that has a critical angle of 68.4º when submerged in water? What is the substance, based on Table 25.1?

(b) What would the critical angle be for this substance in air?

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.

Sign-up for free