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

Question: Suppose the prism of Fig. 33-53 has apex angle ϕ=60oand index of refractionn=1.60 . (a) What is the smallest angle of incidenceθ for which a ray can enter the left face of the prism and exit the right face? (b) What angle of incidenceθ is required for the ray to exit the prism with an identical angleθ for its refraction, as it does in Fig. 33-53?

Figure:

Short Answer

Expert verified
  1. The smallest angle of incidence for which a ray can enter the left face of the prism and exit the right face is 35.6o.
  2. The angle of incidence required for the ray to exit the prism with an identical angle of θ for its refraction is 53.1o.

Step by step solution

01

Given

The apex angle of the prism is, ϕ=60.0o

The Refractive index is, n=1.60

02

Understanding the concept

Using the concept of total internal reflection and critical angle, we can determine the results.

Formula:

Critical angle,θc=sin-1n2n1

03

(a) Calculate The smallest angle of incidence for which a ray can enter the left face of the prism and exit the right face

If the original ray is incident at point A, the prism vertex is at point B, and the point where the interior ray strikes the right surface of the prism is the point C.

The angle between the line AB and the interior ray is βthe angle between the line AC and the interior ray is α.

When the incident ray is at the minimum angle for which light can exit the prism, the light exits along the second face. That is, the angle of refraction at the second face is 90and the angle of incidence there for the interior ray is the critical angle for the total internal reflection.

Let θ1 be the angle of incidence for the original incident ray; let be the angle of refraction at the first face, and let be the angle of incidence at the second face.

The law of refraction applied to point C, yields nsinθ3=1

sin3=1n=11.60=0.625

The interior angles of the triangle ABC must sum to 180.

So, α+β=180

Now,

α=90-θ3=51.32

So,

β=120-51.32=69.68o

Thus,

θ2=90-β=21.32

The law of refraction, applied to point A , yields

.sinθ1=nsinθ2sinθ1=1.60×sin21.32sinθ1=0.5817θ1=sin-10.5817θ1'=35.6

Thus, θ1=35.6

04

(b) Calculate the angle of incidence requires for the ray to exit the prism with an identical angleθfor its refraction.

We apply the law of refraction to point C.

Since the angle of refraction, there is the same as the angle of incidence at A,

nsinθ3=sinθ

Now, α+β=120o

α=90o-θ3,

And β=90-θ2, as before.

This means θ2+θ3=60

Thus, the law of refraction leads to

localid="1662998883723" sinθ1=nsin60-θ2sinθ1=nsin60cosθ2-nsin60cosθ2Fromthetrigonometryidentitysin(A-B)=sinAcosB-cosAsinB

Next, we apply the law of refraction to point A.

sinθ1=nsinθ2sinθ2=sinθ1n

Hence,

cosθ2=1-sin2θ2=1-1n2sin2θ1

Now,

sinθ1=n1sin6001-1n2sin2θ1-cos60sinθ1

1+cos60sinθ1=sin60n2-sin2θ1

Squaring on both sides and solving it for θ1 we get

sinθ1=nsin60o(1+cos60+sin2600sinθ1=1.60×sin60o(1+cos60+sin2600

That gives

sinθ1=0.80

And hence,

θ1=53.1

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

In Fig. 33-47a, a light ray in an underlying material is incident at an angleon a boundary with water, and some of the light refracts into the water. There are two choices of the underlying material. For each, the angle of refractionversus the incident angleis given in Fig. 33-47b.The horizontal axis scale is set byθ1s=90°.Without calculation, determine whether the index of refraction of

(a) material1 and

(b) material2 is greater or less than the index of water(n=1.33).What is the index of refraction of

(c) material 1 And

(d) material 2?

A point source of light emits isotopically with a power of 200 W . What is the force due to the light on a totally absorbing sphere of radius 2.0 cm at a distance of 20 mfrom the source?

In Fig. 33-42, unpolarized light is sent into a system of three polarizing sheets, which transmitsthe initial light intensity. The polarizing directions of the first and third sheets are at anglesθ1=0°andθ3=90°.What are the

(a) smaller and

(b) larger possible values of angleθ2(<90°)for the polarizing direction of sheet 2?

1 A helium–neon laser,radiating at, 632.8nmhas a power output of 3.0mW. The beam diverges (spreads) at angleθ=0.17mrad(Fig. 33-72). (a) What is the intensity of the beamfrom the laser? (b) What is the power of a point source providing that intensity at that distance?

A certain helium–neon laser emits red light in a narrow band of wavelengths centered at632.8nm and with a “wavelength width” (such as on the scale of Fig. 33-1) of0.0100nm. What is the corresponding “frequency width” for the emission?

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