EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
1st Edition
ISBN: 9781337684668
Author: Katz
Publisher: VST
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Question
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Chapter 38, Problem 125PQ

(a)

To determine

The proof for the statement, “the minimum deviation satisfies the equation n=sin(αmin+β2)sin(β2).”

(a)

Expert Solution
Check Mark

Answer to Problem 125PQ

The proof for the statement, “the minimum deviation satisfies the equation n=sin(αmin+β2)sin(β2)” is as stated 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 angle of deviation of the given prism.

    α=θ1+θ4β                                                                                                    (II)

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

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

    θ2+θ3=β                                                                                                         (III)

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

For minimum deviation, refracted angle at both faces must be equals.

    θ2=θ3

For minimum deviation, incidence angle at both faces must be equal.

    θ1=θ4

Substitute θ2 for θ3 in equation (III) to find θ2.

    θ2+θ2=βθ2=β2

Substitute θ1 for θ4 , αmin for α in equation (II) to find θ1.

    αmin=θ1+θ1β2θ1=α+βθ1=αmin+β2

Here, αmin is the minimum deviation angle of the prism.

Further, substitute αmin+β2 for θ1 and β2 for θ2 in equation (I) to find the refractive index.

    sin(αmin+β2)=nsin(β2)

Therefore, the refractive index is,

    n=sin(αmin+β2)sin(β2)                                                                                               (V)

Conclusion:

Therefore, refractive index is sin(αmin+β2)sin(β2).

(b)

To determine

The minimum deviation angle.

(b)

Expert Solution
Check Mark

Answer to Problem 125PQ

The minimum deviation angle is αmin=β(n1).

Explanation of Solution

Use the approximation method for angle.

    sinββ

Conclusion:

Substitute αmin+β2 for sin(αmin+β2) and β2 for sin(β2) in equation (V) to find the minimum deviation.

    n=(αmin+β2)β2nβ=αmin+βαmin=β(n1)

Therefore, minimum deviation angle is αmin=β(n1).

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Chapter 38 Solutions

EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC

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