Try the fastest way to create flashcards
Question

# A beam of light traveling in an unknown material encounters a boundary with flint glass (n = 1.66). If the critical angle for total internal reflection is $32^{\circ}$, what is the index of refraction of the unknown material?

Solution

Verified
Step 1
1 of 2

Snell Law states:

$n_1\sin\theta_1=n_2\sin\theta_2$

Where $n_1$ and $n_2$ are the indexes of refraction respectively of the medium 1 and 2, $\theta_1$ is the angle of the ray to the normal in medium 1 and $\theta_2$ is the angle of the ray to the normal in medium 2.

Total internal reflection can happen when light is trying to enter a medium with a lower index of refraction, we call the first angle at which we observe this phenomenon the critical angle $\theta_c$, at this angle, the light travels parallel to the surface separating the two mediums yielding a $90^o$ degree angle with the normal.

We can then assert the following:

$n_1\sin\theta_c=n_2\sin 90^o$

$n_1=\frac{n_2}{\sin\theta_c}=\frac{1.66}{\sin 32^o}=\boxed{3.13}$

## Recommended textbook solutions

#### Physics

1st EditionISBN: 9780131371156Walker
3,630 solutions

#### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th EditionISBN: 9780133942651 (8 more)Randall D. Knight
3,508 solutions

#### Mathematical Methods in the Physical Sciences

3rd EditionISBN: 9780471198260 (1 more)Mary L. Boas
3,355 solutions

#### Fundamentals of Physics

10th EditionISBN: 9781118230718 (3 more)David Halliday, Jearl Walker, Robert Resnick
8,971 solutions