Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 23, Problem 18P
To determine

The expression for the angle of deviation δ in term of β and n and to show that δ is proportional to the angle β of the prism.

Expert Solution & Answer
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Answer to Problem 18P

The expression for the angle of deviation is δ=β(n1), and the value of δ is proportional to the angle β of the prism.

Explanation of Solution

The schematic ray diagram involving the prism is shown in Figure 1. The refractive index of the medium outside the prism is n1, and that of the material of the prism is n2. The angles α, β,γ, δ, θ1, θ2, θ3, and θ4 are marked in the Figure 1.

Physics, Chapter 23, Problem 18P

From the Figure 1, the following relations can be deduced.

β+2α=180° (I)

θ1=90°α (II)

Solve equation (I) for α.

α=180°β2=90°β2 (III)

Use equation (III) in (II).

θ1=90°(90°β2)=β2 (IV)

Apply Snell’s law to the boundary where light enters the prism.

n1sinθ1=n2sinθ2 (V)

Solve equation (V) for sinθ2.

sinθ2=n1n2sinθ1 (VI)

Use equation (IV) in (VI).

sinθ2=n1n2sinβ2 (VII)

Use the approximation sinθθ in equation (VII) since β is very small.

sinθ2n1n2β2 (VIII)

Use the approximation sinθθ in equation (VIII) since θ2 is also small.

θ2n1n2β2 (IX)

Use n1=1 (refractive index of air), and n2=n to simplify equation (IX).

θ2=β2n (X)

In the triangle formed by the ray crossing the prism and the two surfaces of the prism, the angles are (90°θ2), β, and (90°θ3). As interior angles of a triangle, the sum of these angles must be 180°.

(90°θ2)+β+(90°θ3)=180° (XI)

Solve equation (XI) for θ3.

θ3=βθ2 (XII)

Use equation (X) in (XII).

θ3=ββ2n=β(112n) (XIII)

Apply Snell’s law to the boundary where light leaves the prism.

n1sinθ=n2sinθ3 (XIV)

Use n1=1 (refractive index of air), and n2=n to simplify equation (XIV).

sinθ4=nsinθ3 (XV)

Use equation (XIII) in (XV).

sinθ4=nsin[β(112n)] (XVI)

Since β(112n)<β, it is also small. Moreover, θ4 is small as well. Thus the approximation sinθθ can be applied to equation (XVI).

θ4nβ(112n)=β(n12) (XVII)

Write the expression for the angle of deviation δ from the figure.

δ=θ4θ1 (XVIII)

Use equation (XVII) and (IV) in (XVIII).

δ=β(n12)β2=β(n1212)=β(n1)

This indicates that, the angle of deviation δ is proportional to the angle β of the prism.

Conclusion:

Therefore, the expression for the angle of deviation is δ=β(n1), and the value of δ is proportional to the angle β of the prism.

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