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# Rainbow. above figure (a), shows a light ray entering and then leaving a falling, spherical raindrop after one internal reflection (see above figure (b)). The final direction of travel is deviated (turned) from the initial direction of travel by angular deviation $$\theta _{dev}$$. (a) Show that $$\theta _{dev}$$ is $$\theta _{dev}=180^{0}+2\theta _{i}-4\theta _{r}$$,where $$\theta _{i}$$ is the angle of incidence of the ray on the drop and $$\theta _{r}$$ is the angle of refraction of the ray within the drop. (b) Using Snells law, substitute for $$\theta _{r}$$ in terms of $$\theta _{i}$$ and the index of refraction $$n$$ of the water. Then, on a graphing calculator or with a computer graphing package, graph $$\theta _{dev}$$ versus $$\theta _{i}$$ for the range of possible $$\theta _{i}$$ values and for $$n=1.331$$ for red light (at one end of the visible spectrum) and $$n=1.333$$ for blue light (at the other end). The red-light curve and the blue-light curve have different minima, which means that there is a different angle of minimum deviation for each color. The light of any given color that leaves the drop at that colors angle of minimum deviation is especially bright because rays bunch up at that angle. Thus, the bright red light leaves the drop at one angle and the bright blue light leaves it at another angle. Determine the angle of minimum deviation from the $$\theta _{dev}$$ curve for (c) red light and (d) blue light. (e) If these colors form the inner and outer edges of a rainbow (above figure (b)), what is the angular width of the rainbow?

Solution
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#### (a) The first contribution to the overall deviation is at the first refraction: $$\delta \theta _{1}=\theta _{i}-\theta _{r}$$. The next contribution to the overall deviation is the reflection. Noting that the angle between the ray right before reflection and the axis normal to the back surface of the sphere is equal to $$\theta _{r}$$, and recalling the law of reflection, we conclude that the angle by which the ray turns (comparing the direction of propagation before and after the reflection) is $$\delta \theta _{2}=180^{0}-2\theta _{r}$$. The final contribution is the refraction suffered by the ray upon leaving the sphere: $$\delta \theta _{3}=\theta _{i}-\theta _{r}$$ again. Therefore,$$\theta _{dev}=\delta \theta _{1}+\delta \theta _{2}+\delta \theta _{3}=180^{0}+2\theta _{i}-4\theta _{r}$$.(b) We substitute $$\theta _{r}=\sin ^{-1}\left ( \frac{1}{n}\sin \theta _{i} \right )$$ into the expression derived in part (a), using the two given values for $$n$$. The higher curve is for the blue light.(c) We can expand the graph and try to estimate the minimum, or search for it with a more sophisticated numerical procedure. We find that the $$\theta _{dev}$$ minimum for red light is $$137.63° ≈ 137.6°$$, and this occurs at $$\theta _{i}=59.52^{0}$$.(d) For blue light, we find that the $$\theta _{dev}$$ minimum is $$139.35° ≈ 139.4°$$, and this occurs at $$\theta _{i}=59.52^{0}$$.e) The difference in $$\theta _{dev}$$ in the previous two parts is $$1.72^{0}$$.

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