The Angle of Deviation. When a light ray hits a raindrop, the light undergoes a sequence of reflections inside the water droplet. The total rotation that the light ray undergoes is given by

Question

The input x measures the angle (in radians) that the light ray forms with the surface of the spherical raindrop so the domain of D is The output, D(X), is also an angle in radians. The constant is called the index of refraction and depends on the color of the light.

Find the minimum value of D when Using trig identities, it is possible to write your answer in the form arccos (r) for some rational number r. Your solution should show the derivative, your calculation of the critical number, and the final maximum value in the form arccos (r)

Answer 1

To solve this problem no physics is required.

In general, an extremum of any continuous differentiable function on a closed domain

occurs either at the boundary of its domain or at a critical point.

So we just need to find any critical points and compare the value of D(x)

at the critical points with its values at 0 and π.

There is only one tricky thing here and that is the range of sin^-1(x).

The most common convention is to make the range [-π/2, +π/2].

And I assume that is the convention in the definition of D(x).

Therefore the usually expected identity

sin^-1(sin(x)) = x

holds only when when the domain is [-π/2, +π/2].

Since your domain is [0, π] that identity does not hold.

I will point out below where that plays a role.

The reason for choosing the range of sin^-1(x) to be [-π/2, +π/2] is to

make its derivative simply

d/dx(sin^-1(x)) = 1/√(1-x²)

Then

D'(x) = 2 - 4(cos(x)/k)/√(1 - sin²(x)/k²) = 2 - 4cos(x)/√(k²- sin²(x))

Setting

D'(x) = 0

2cos(x)/√(k²- sin²(x)) = 1

2cos(x) = √(k²- sin²(x))

Square both sides, but need to be careful.

In general, the equation resulting from squaring may have more solutions than the original.

See below.

4cos²(x) = k² - sin²(x)

4cos²(x) = k² - 1 + cos²(x)

3cos²(x) = k² - 1

cos(x) = ±√((k²-1)/3)

As mentioned above, it is possible that one of the two solutions is not a solution

of the original D'(x) = 0.

And in fact, we need to reject the negative value for cos(x).

Here is the place where the range of sin^-1(x) plays a role.

By making it [-π/2, +π/2], the derivative of sin^-1(x) > 0 and hence the positive value of the square root is taken.

Thus

x = arccos(√((k²-1)/3))

Substituting k = √(5/2)

x = arccos(√((5/2-1)/3)) = arccos(√(1/2)) = π/4

So we have a critical point at π/4, but we need to figure out whether it is

maximum, minimum, or neither.

This is particular important in this problem, whose statement asks you to

find minimum in one sentence and maximum in another one.

So we need to compare the value of D(x) at the domain boundary with its value

at the critical point.

D(0) = π

D(π) = 3π

D(π/4) = π + π/2 - 4(sin^-1(√(2/5)√(1/2)))

= π + π/2 - 4(sin^-1(√(1/5)

= π + π/2 - 1.8549 < π

Therefore, the function D(x) attains its minimum at x = π/4 and its maximum at x = π.