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Chapter 23: Problem 100

The angle of deviation through a triangular prism is defined as the angle between the incident ray and the emerging ray (angle \(\delta\) ). It can be shown that when the angle of incidence \(i\) is equal to the angle of refraction \(r^{\prime}\) for the emerging ray, the angle of deviation is at a minimum. Show that the minimum deviation angle \(\left(\delta_{\min }=D\right)\) is related to the prism angle \(A\) and the index of refraction \(n,\) by $$ n=\frac{\sin \frac{1}{2}(A+D)}{\sin \frac{1}{2} A} $$ [Hint: For an isosceles triangular prism, the minimum angle of deviation occurs when the ray inside the prism is parallel to the base, as shown in the figure.]

Short Answer

Expert verified
Question: Prove that the minimum angle of deviation for a given triangular prism is related to the prism angle and the index of refraction using Snell's law and geometrical optics, based on the given formula \(n=\frac{\sin \frac{1}{2}(A+D)}{\sin \frac{1}{2} A}\).

Step by step solution

01

Apply Snell's law at the first face of the triangular prism

Using Snell's law, we can relate the incident angle, index of refraction, and the angle of refraction for the first face of the prism, $$ n = \frac{\sin i}{\sin r} $$
02

Apply Snell's law at the second face of the triangular prism

For the second face of the prism, again applying Snell's law, we have, $$ \frac{1}{n} = \frac{\sin r^{\prime}}{\sin i^{\prime}} $$
03

Use given condition \(i = r^{\prime}\)

The given condition states that the angle of incidence \(i\) is equal to the angle of refraction \(r^{\prime}\) for the emerging ray, $$ i = r^{\prime} $$
04

Find the relationship between \(r\), \(A\), and \(i^{\prime}\)

Since both the incident ray and the emerging ray are parallel to the base of the isosceles triangular prism, we use the geometry of the prism to find the relationship, $$ r + A = 180^\circ - i^{\prime} $$
05

Use the relationships from Steps 3 and 4 to find \(i^{\prime}\)

Combining the equations from Steps 3 and 4, we get, $$ r = i^{\prime} - A $$
06

Calculate \(\sin i^{\prime}\) and \(\sin r\) from Step 5

Using the sine function, we get the relationship between \(\sin i^{\prime}\) and \(\sin r\), $$ \sin r = \sin (i^{\prime} - A) $$
07

Find the relationship between \(i\) and \(D\)

Now, we use the geometry of the ray path to find the relationship between the incident angle and the minimum angle of deviation, $$ D = i + i^{\prime} - A $$
08

Use Snell's law to find a relationship between \(D\) and \(A\)

Substituting \(\sin r\) and \(\sin i^{\prime}\) from Steps 1, 2, and 6 into Snell's law, we get, $$ \frac{\sin i}{\sin (i^{\prime} - A)} = n = \frac{\sin (i^{\prime} - i)}{\sin (i^{\prime} - A)} $$ Now, using the relationship between \(i\) and \(D\) from Step 7, we can rewrite this as, $$ n = \frac{\sin \frac{1}{2}(D - A)}{\sin (-\frac{1}{2}A)} $$ Multiplying both sides by -1, we finally obtain the desired relationship, $$ n = \frac{\sin \frac{1}{2}(A + D)}{\sin \frac{1}{2} A} $$

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