What is the Minimum Value of Refractive Index for Total Internal Reflection?

AI Thread Summary
The discussion revolves around calculating the minimum refractive index required for total internal reflection using Snell's Law. The initial calculation yields a refractive index of 1.41 when applying the formula with a 45-degree angle. It is clarified that total reflectance occurs when n*sin(45) is greater than or equal to 1, leading to the conclusion that n must be at least 1/sqrt(2). The conversation emphasizes that while the calculated value is correct, total internal reflection can occur at higher refractive indices as well. Understanding the concept of total internal reflection is crucial for further insights.
SirKhairin
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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
 

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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
 
I still can't get it
 
The vocabulary phrase you want to look up is "total internal reflection."
 
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