Solutions to Part 17 (Refraction)

Solutions (problems given first):

Other problems:

1) For the same set up and experiment, obtain the values the student should get for the angles i2 and r2 for the emergent ray if his results are to be consistent.

To obtain the same index of refraction: (1.73), the student will apply Snell’s law in the form:

sin i2/ sin r2 = 1.73 = = [3]^½

This means: i2 = 60 degrees, and r2 = 30 degrees.


2)If the speed of light in air is 300,000 km/sec what is the speed in a substance with refractive index 1.74?

We have:

n1 v1 = n2 v2

where n1 = 1.74 and v1 is unknown with v2 = 300,000 km/sec and n2 (refractive index of air) = 1.0

then: v1 = (n2 v2)/ n1 = (300,000 km/sec)/ 1.74 = 172, 400 km/sec


3)Light in air has a wavelength of 0.0000589 cm. What would the wavelength be in water for which n(w) = 4/3.


First note that: v = fL for any light wave where f = frequency and L = wavelength

Then, on comparing conditions in two media one will have:

v1 = f1 L1 and v2 = f2 L2

Now: n1 v1 = n2 v2

So that on substituting (for v1 and v2, speeds in the different media):

n1( f1 L1) = n2 (f2 L2)

In air we know n1 =1 so v1 is speed in air and L1 is wavelength in air ( = 0.0000589 cm)

Then, on simplifying:

L1 = L2 n2 where n2 = 4/3

Then: L2 (wavelength in water) = L1/ n1 = (0.0000589 cm)/ 4/3

L2 = 4.4 x 10^-5 cm



4) When the angle i1 is equal to π/2, i2 is equal to what is called the critical angle. For this case: sin(π/2) = 1 so: 1 = n sin (i2).

If the index of refraction of a piece of extra dense flint glass is 1.60, would the critical angle be greater or less than 60 degrees?

The sketch of the geometry of the situation is shown in the accompanying diagram.

And we know: 1 = (1.60) (sin i2) so:

1/(1.60) = sin(i2) = 0.625

Which provides an angle < 60 degrees (since sin(60) = 0.866)