It's Critical that you get the right angle!

Solutions to the AP Problems:
79B6.
a) Remember that you MUST use the angles relative to the normal vector - n1 sin(t1) = n2 sin (t2)
n2 = 1.327
n = c/v so v = 2.26 x 10⁸ m/s

b) If you try to use Snell's law at point Q, you will find you end up trying to take the inverse sine of a number greater than 1 - not possible. This means the ray will NOT refract, but instead reflect. This is total internal reflection. You can justify the fact that it reflects either by calculating the critical angle and showing that the 53 degrees is greater than the critical angle (49 degrees), or by showing that when refraction does not occur, reflection does.

c) If the lens was made of plastic in air, it would be a converging lens. Since it is the reverse, it will be a diverging lens, and the rays will spread out.

88B5.
a) The ray will be refracted slightly downwards from the horizontal dotted line.

b)
The ray will pass directly through the left interface because the incident angle is 0. At the right interface, the angle from the normal is 37 degrees. By Snell's law, the refracted ray will have an angle of 64.5 degrees.

Be careful though, as the question asks for the angle from the horizontal. To get this, you must subtract the 37 degrees from the 64.5 degrees, giving a final answer of 27.5 degrees.

c) Here you must assume that the incident angle of 37 degrees is the critical angle.

By Snell's law, theta_1 = 37, n2 = 1.0 (for air), and theta_2 = 90 since we are assuming total internal reflection occurs. Solving, n1 = 1.667

d) Here the light ray will again be refracted below the horizontal line, but the angle should not be as great as it was for part (a).

e) Using Snell's law again with n1 = 1.667, theta_1 = 37 degrees, and n2 = 1.33, we can solve for theta_2 = 49.0 degrees. You must again subtract 37 degrees from this value to obtain the angle from the horizontal of 12 degrees below.

93B4.
a) v = c/n = 1.875 x 10⁸ m/s
b) There are two ways to do this - the fast way is to remember the equation from Monday that the wavelength in the material lambda' = lambda/n where lambda is the wavelength in a vacuum. 700 nm/1.5 = 467 nm.

We can also use c = f*lambda to find the frequency of the red light in a vacuum:
f = 3E8 m/s/(700 E-9 m) = 4.28 x 10¹⁴ Hz

The frequency stays the same when a wave changes from one material to another, so this is also the frequency in the glass. We also know the speed of red light in the glass from part a.

lambda = v/f = 1.875 x 10⁸ m/s/(4.28 x 10¹⁴ Hz) = 438 nm (either answer would be acceptable)

c) Frequency can be found the same as before - make SURE before you use the v = f*lambda equation, you are using the speed and wavelength for the same medium!

f = 3E8 m/s/(700 E-9 m) = 4.28 x 10¹⁴ Hz

d) The key here is to notice that the incident rays have the SAME initial angle hitting the right side of the prism (30 degrees), as well as the same n2 = 1 (for air.) The difference is in n1.

From Snell's Law:
n1 sin(theta_1) = n2*sin(theta_2)

sin(theta_2) = n1*sin(theta_1)

Since n1 is greater for the blue light, theta_2 will be greater for the blue light. This means that the blue light will bend further away relative to the normal than the red.

e) In this case, it is a similar situation when the rays reach the right surface (the rays again do not refract at the left surface because the incident angle is zero.) The issue here, however, is that n1 is the same for both (n1 = 1), and theta_1 is again 30 degrees.

Snell's law again:

n1 sin(theta_1) = n2*sin(theta_2)

sin(theta_2) = 1*sin(theta_1)/n1

In this case, since n1 is greater for blue light, it will refract less
than with the red light. Thus, the blue light will bend above the red light, with both rays bending towards the normal.

#3.
b) Using the thin lens equation, 1/di = 1/f - 1/do, so di = 30 cm. This should roughly match your answer for part (a).

c) hi/ho = -di/do so hi = 2*5cm = 10 cm

d) This can be a tricky one. If you use what we came up with today, you will find that the image from the first lens is on the right side of the second lens. You can use the rules as we always do, except that the primary focus of the second lens is now on the same side as the object, so the lens will act like a diverging lens.

The slick way to get this is to notice that Ray 2 leaves the first lens parallel to the axis when it enters the second lens. This ray enters lens 2 parallel, so it is brought to the primary focus of lens 2. Ray 1, after passing through lens 1, passes through the center of lens 2, which means it will pass through undeflected. This will result in these two rays converging at a single point to the right of lens 2. This is the location of the image, roughly 7 cm to the right of lens 2.

e) The image is inverted and smaller than the original object.