# Refractive Index question

1. Jan 24, 2010

### SirKhairin

1. The problem statement, all variables and given/known data

2. The attempt at a solution

At first what I did was using the Snell's Law formula $$n_1 \sin{\theta _1}=n_2 \sin{\theta _2}$$

then I did this, $$n\sin{\ 45}=1 \sin{\ 90}$$

so this is the minimum value that I got n=1.41

I'm not sure if my answer is correct or not. Also how do you justify that particular n value is minimum

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2. Jan 24, 2010

### ehild

Total reflectance occurs here when $n\sin(45)\geq 1$, so $n\geq 1/\sqrt{2}$. Your value is minimum as total reflectance holds for higher refractive index values, too.

ehild

3. Jan 25, 2010

### SirKhairin

I still can't get it

4. Jan 25, 2010

### mikelepore

The vocabulary phrase you want to look up is "total internal reflection."