EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
1st Edition
ISBN: 9780100546714
Author: Katz
Publisher: YUZU
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Chapter 38, Problem 124PQ

(a)

To determine

The proof that total deviation of a ray from its original path is θ1β+sin1(sinβn2sinθ12cosβsinθ1).

(a)

Expert Solution
Check Mark

Answer to Problem 124PQ

The total deviation of a ray from its original path is

θ1β+sin1(sinβn2sinθ12cosβsinθ1) and is proved below.

Explanation of Solution

Write the expression for refraction law at left side of the given prism.

    sinθ1=nsinθ2 (I)

Here, θ1 is incident angle of ray coming from left at left surface, θ2 is the refracted angle at left face of the prism and n is refractive index of prism.

Write the expression for refraction law at right side of the given prism.

    nsinθ3=sinθ4                                                                                                    (II)

Here, θ3 is incident angle of ray coming from left at right surface, θ4 is the refracted angle at left face of the prism

Write the expression angle of deviation of the given prism.

    α=θ1+θ4β                                                                                                  (III)

Here, α is angle of deviation and β is apex angle of the prism.

Write the relation among various angle from the geometry of the prism.

    θ2+θ3=βθ3=βθ2

Substitute βθ2 for θ3 in equation (II) to find θ4.

    nsin(βθ2)=sinθ4θ4=sin1(nsin(βθ2))

Further solve this for θ4.

    θ4=sin1(nsinβcosθ2ncosβsinθ2)                                                           (IV)

Use general trigonometric relation for conversion.

    sin2θ+cos2θ=1

Substitute 1sin2θ2 for cosθ2 in equation (IV) to find θ4.

    θ4=sin1(nsinβ1sin2θ2ncosβsinθ2)

Substitute sinθ1n for sinθ2 in above equation to find θ4.

    θ4=sin1(nsinβ1(sinθ1n)2ncosβ(sinθ1n))=sin1(sinβn2sinθ12cosβsinθ1)

Substitute sin1(sinβn2sinθ12cosβsinθ1) for θ4 in equation (III) to find angle of deviation.

    α=θ1+sin1(sinβn2sinθ12cosβsinθ1)β=θ1β+sin1(sinβn2sinθ12cosβsinθ1)

Conclusion:

Therefore, it is proved that the angle of deviation is θ1β+sin1(sinβn2sinθ12cosβsinθ1).

(b)

To determine

The plot of the angle of deviation versus the angle of incidence for n=1.5 and apex angle 60.0°

(b)

Expert Solution
Check Mark

Answer to Problem 124PQ

The graph is following.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 38, Problem 124PQ , additional homework tip  1

Explanation of Solution

Write the expression for the angle of deviation of the given prism.

    α=θ1β+sin1(sinβn2sinθ12cosβsinθ1)                                             (V)

Conclusion:

Substitute 1.5 for n, 60.0° for β in above equation to find α

    α=θ160+sin1(sin(60.0°)(1.5)2sinθ12cos(60.0°)sinθ1)=θ160+sin1(0.872.25sinθ12(0.5)sinθ1)

Deviation varies linearly, decreases and then increases approximately with incidence angle θ1.

Therefore, the plot of the angle of deviation versus the angle of incidence is as follows.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 38, Problem 124PQ , additional homework tip  2

Figure-(1)

(c)

To determine

The incidence angle for which the derivation angle is minimum.

(c)

Expert Solution
Check Mark

Answer to Problem 124PQ

The incidence angle for which deviation is minimum is 48° if the light deviated by this angle passes symmetrically through the prism.

Explanation of Solution

The minimum deviation occurs when the incident and the refracted ray are identical and make equal angles to the normal of the prism.

The angle exhibits a minimum at 48°, if the light deviated by this angle passes symmetrically through the prism. So the angle of incidence to the prism and angle of departure are equal.

Conclusion:

The minimum angle of deviation depends on the refractive index for different wavelength, the refractive index is different.

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Chapter 38 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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