/*! 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} Q83P A ray of white light traveling t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A ray of white light traveling through fused quartz is incident at a quartz–air interface at angleθ1.Assume that the index of refraction of quartz is n=1.456at the red end of the visible range andn=1.470 at the blue end.Ifθ1 is(a) θ1=42.00°, (b),θ1=43.10°and (c),θ1=44.00°is the refracted light white, white dominated by the red end of the visible range, or white dominated by the blue end of the visible range, or is there no refracted light?

Short Answer

Expert verified

(a) The refracted light for the given incidence angle42.00° will be white light.

(b) The refracted light for the given incidence angle 43.10°will be white dominated by the red end of the visible end.

(c) There is no refracted light for the given incidence angle.44.00°

Step by step solution

01

The given data:

  • The index of refraction of quartz at the red end of the visible range is,n=1.456.
  • The index of refraction of quartz at the blue end of the visible range is,n=1.470 .
02

Understanding the concept of refraction: 

Use the concept of refraction of light. Calculate the critical angle for the quartz at both the red end of the visible range and at the blue end. So, you can compare the given incidence angle with the critical angle and can determine the nature of refracted light.

Formula:

The critical angle of a refracted light with the incident light,

θc=sin−1(n2n1)….. (1)

03

a) Calculation of the refracted light for angle  : 42° 

Here, the ray of white light is incident at a quartz-air interface. Therefore,

The index of refraction of air, n2=1

The index of refraction, n1=n

Thus, using this value in equation (1), you can get the equation of critical angle as follows:

θc=sin−1(1n) ….. (2)

Thus, the critical angle for the red end of the visible range is given using the given data in equation (2) as:

(θc)red=sin−1(11.456)=43.38°

Similarly, the critical angle for the blue end of the visible range is given using the given data in equation (a) as:

(θc)blue=sin−1(11.470)=42.86°

Now, considering the incidence angleθ1as42.00° , you get that the incident angle is less than both critical angle for red end and critical angle for blue end.

So, all types of rays can be refracted. Hence, the refracted light will be white light.

04

(b) Calculation of the refracted light for angle  :43.10°

Now, considering the incidence angleθ1as 43.10°, you get that the incident angle is greater than the critical angle for the blue end of the visible range and less than the red end of the visible range.

Hence, the refracted light will be white-dominated by the red end of the visible end.

05

(c) Calculation of the refracted light for angle  :

Now, considering incidence angleθ1as 44.00°, you get that the incident angle is less than both the critical angle for the red end and the critical angle for the blue end.

Hence, no visible light can be refracted, and you can say that there is no refracted light.

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

In Fig. 33-41, unpolarized light is sent into a system of two polarizing sheets. The anglesof the polarizing directions of the sheets are measured counterclockwise from the positive direction of the y-axis (they are not drawn to scale in the figure). The angle θ1 is fixed but the angle θ2can be varied. Figure 33-45 gives the intensity of the light emerging from sheet 2 as a function of θ2 . (The scale of the intensity axis is not indicated.) What percentage of the light’s initial intensity is transmitted by the two-sheet system when?

a.

b.

Two polarizing sheets, one directly above the other, transmit p% of the initially unpolarized light that is perpendicularly incident on the top sheet. What is the angle between the polarizing directions of the two sheets?

Question: (a) How long does it take a radio signal to travel 150kmfrom a transmitter to a receiving antenna? (b) We see a full Moon by reflected sunlight. How much earlier did the light that enters our eye leave the Sun? The Earth–Moon and Earth–Sun distances are 3.8x105km and , respectively. (c) What is the round-trip travel time for light between Earth and a spaceship orbiting Saturn, 1.3x109km distant? (d) The Crab nebula, which is about 6500light years (ly) distant, is thought to be the result of a supernova explosion recorded by Chinese astronomers in A.D. 1054. In approximately what year did the explosion actually occur? (When we look into the night sky, we are effectively looking back in time.)

A plane electromagnetic wave traveling in the positive direction of anxaxis in vacuum has componentsEx=Ey=0andEz=(2.0V/m)cos[π×1015s-1t-x/c]. (a) What is the amplitude of the magnetic field component? (b) Parallel to which axis does the magnetic field oscillate? (c) When the electric field component is in the positive direction of thezaxis at a certain pointP, what is the direction of the magnetic field component there?

Start from Eq. 33-11 and 33-17 and show that E(x,t),and B(x,t),the electric and magnetic field components of a plane traveling electromagnetic wave, must satisfy the wave equations

∂2E∂t2=c2∂2E∂x2and∂2B∂t2=c2∂2B∂x2

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.