- #1

Karol

- 1,380

- 22

## Homework Statement

$$\frac{\sin\theta_1}{c_1}=\frac{\sin\theta_2}{c_2}$$

$$\frac{c_1}{c_2}=n_{12}$$

Express ##\theta_2## as a function of ##\theta_1##

Find the largest value of ##\theta_1## for which the expression for ##\theta_2## that you just found is defined (for larger values of ##\theta_1## than this the incoming light will be reflected).

## Homework Equations

Inverse sine: ##y=\sin^{-1}(x)~\rightarrow~\sin(y)=x##

## The Attempt at a Solution

$$\sin ( \theta_2 )=\frac{\sin( \theta_1 ) }{n_{12}}~\rightarrow~\sin^{-1}\left( \frac{\sin( \theta_1 )}{n_{12}} \right)=\theta_2$$

##\theta_2## can be ##\frac{\pi}{2}## at the max:

$$\sin^{-1}\left( \frac{\sin( \theta_1 )}{n_{12}} \right)=\frac{\pi}{2}~~\rightarrow~~\sin\left( \frac{\pi}{2} \right)=\frac{\sin(\theta_1)}{n_{12}}$$

$$\Rightarrow~\sin(\theta_1)=n_{12}\sin\left( \frac{\pi}{2} \right)=n_{12}$$

$$\theta_1<\arcsin(n_{12})$$

I didn't use at all the definition of inverse function in the second question, i feel what i have done isn't what it's meant from the chapter of inverse functions