(III) A beam of light enters the end of an optic fiber as shown in Fig. 32–59. (a) Show that we can guarantee total internal reflection at the side surface of the material (at point A), if the index of refraction is greater than about 1.42. In other words, regardless of the angle α , the light beam reflects back into the material at point A, assuming air outside. (b) What if the fiber is immersed in water?

(II) In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, 1.3 m above the water level, onto the surface of the water at a point 2.8 m from his foot at the edge of the pool (Fig. 32–53). Where does the spot of light hit the bottom of the pool which is 2.1 m deep? Measure from the bottom of the wall beneath his foot.

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(II) (a) What is the minimum index of refraction for a glass or plastic prism to be used in binoculars (Fig. 32–34) so that total internal reflection occurs at 45°? (b) Will binoculars work if their prisms (assume n = 1.58) are immersed in water? (c) What minimum n is needed if the prisms are immersed in water?

(II) An aquarium filled with water has flat glass sides whose index of refraction is 1.51. A beam of light from outside the aquarium strikes the glass at a 43.5° angle to the perpendicular (Fig. 32–52). What is the angle of this light ray when it enters (a) the glass, and then (b) the water? (c) What would be the refracted angle if the ray entered the water directly?

(II) For any type of wave that reaches a boundary beyond which its speed is increased, there is a maximum incident angle if there is to be a transmitted refracted wave. This maximum incident angle θ_iM corresponds to an angle of refraction equal to 90°. If θᵢ > θ_iM, all the wave is reflected at the boundary and none is refracted, because this would correspond to sin θᵣ > 1 (where is the angle θᵣ of refraction), which is impossible.

(a) Find a formula for θ_iM using the law of refraction, Eq. 15–19.

(III) A light ray is incident on a flat piece of glass with index of refraction n as in Fig. 32–24. Show that if the incident angle θ is small, the emerging ray is displaced a distance d = tθ(n - 1)/n , where t is the thickness of the glass, θ is in radians, and d is the perpendicular distance between the incident ray and the (dashed) line of the emerging ray (Fig. 32–24).

When light passes through a prism, the angle that the refracted ray makes relative to the incident ray is called the deviation angle δ, Fig. 32–64. Show that this angle is a minimum when the ray passes through the prism symmetrically, perpendicular to the bisector of the apex angle Φ, and show that the minimum deviation angle, δ_m, is related to the prism’s index of refraction n by

$n=\frac{\sin \frac{1}{2}(\varphi +{\delta }_m)}{\sin \varphi /2}.$

[Hint: For θ in radians, $\left(d/d\theta \right)\left({\sin }^{-1}\theta \right)=1/\sqrt{1-{\theta }^2}$.]