Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
5th Edition
ISBN: 9781305586871
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 25, Problem 25P

A ray of light strikes the midpoint of one face of an equiangular (60°−60°−60°) glass prism (n = 1.5) at an angle of incidence of 30°. (a) Trace the path of the light ray through the glass and find the angles of incidence and refraction at each surface. (b) If a small fraction of light is also reflected at each surface, what are the angles of reflection at the surfaces?

(a)

Expert Solution
Check Mark
To determine

The trace of path of the light ray through the glass and the angles of incidence and refraction at each surface.

Answer to Problem 25P

The angles of incidence and refraction at surface 1 are 30° and 19.47° respectively and the angles of incidence and refraction at surface 2 are 40.53° and 77.1° respectively and the traced path of the light ray through the glass is shown below.

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), Chapter 25, Problem 25P , additional homework tip  1

Explanation of Solution

Introduction:

When a ray incidents on the glass then the rays get refracted through angle of refraction r1 at surface 1 and then refracted through angle of refraction r2 at surface 2 .

Given info: The angle of incidence at surface 1 is 30° and the refractive index of the glass is 1.5 .

The path of the light ray through the glass is shown below,

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card), Chapter 25, Problem 25P , additional homework tip  2

Figure (1)

Write the Snell’s law, for surface 1 ,

n1sini1=n2sinr1

Here,

n1 is the refractive index of air.

i1 is the incidence angle of surface 1 .

n2 is the refractive index of glass.

r1 is the refracted angle of surface 1

Rearrange the above expression for r1 .

r1=sin1[n1n2sini1]

Substitute 1 for n1 , 1.5 for n2 and 30° for i1 in the above equation to find the value of r1 .

r1=sin1[11.5sin30°]=19.47°

Thus the incidence angle at surface 1 is 30° and the refracted angle at surface 1 is 19.47° .

From the figure, the corresponding angles are equal.

QRC=ABC

Substitute 60° for ABC in the above equation.

QRC=60°

From the figure, the alternate angles are equal.

PRC=r1

Substitute 19.47° for r1 in the above equation.

PRC=19.47°

From the figure

i2=QRCPRC

Substitute 60° for QRC and 19.47° for PRC in the above equation.

i2=60°19.47°=40.53°

Thus the angle of incidence at surface 2 is 40.53° .

Write the Snell’s law, for surface 2 ,

n2sini2=n1sinr2

Here,

n1 is the refractive index of air.

i2 is the incidence angle of surface 2 .

n2 is the refractive index of glass.

r2 is the refracted angle of surface 2

Rearrange the above expression for r2 .

r2=sin1[n2n1sini2]

Substitute 1 for n1 , 1.5 for n2 and 40.53° for i2 in the above equation to find the value of r2 .

r2=sin1[1.51sin40.53°]=77.1°

Thus the angle of refraction at surface 2 is 77.1° .

Conclusion:

Therefore, the angles of incidence and refraction at surface 1 are 30° and 19.47° respectively and the angles of incidence and refraction at surface 2 are 40.53° and 77.1° respectively.

(b)

Expert Solution
Check Mark
To determine

The angles of reflection at each surface.

Answer to Problem 25P

The angle of reflection at surface 1 is 30° and the angle of reflection at surface 2 is 40.53° .

Explanation of Solution

Given info: The angle of incidence at surface 1 is 30° and the refractive index of the glass is 1.5 .

From the angle of reflection at surface 1 is,

i1=r1

Here,

i1 is the angle of incidence at surface 1 .

r1 is the angle of reflection at surface 1 .

Rearrange the above expression for r1 .

r1=i1

Substitute 30° for i1 in the above equation to find the value of r1 .

r1=30°

Thus the angle of reflection at surface 1 is 30° .

From the angle of reflection at surface 2 is,

i2=r2

Here,

i2 is the angle of incidence at surface 2 .

r2 is the angle of reflection at surface 2 .

Rearrange the above expression for r2 .

r2=i2

Substitute 40.53° for i2 in the above equation to find the value of r2 .

r2=40.53°

Thus the angle of reflection at surface 2 is 40.53° .

Conclusion:

Therefore, the angle of reflection at surface 1 is 30° and the angle of reflection at surface 2 is 40.53° .

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Chapter 25 Solutions

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)

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