Minimum refractive index for total internal reflection at 45 degrees

[SirKhairin]

Homework Statement

I uploaded the question below.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

 

[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

 

[SirKhairin]

I still can't get it

 

[mikelepore]

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