Refractive Index at Critical Angle: Find the Solution

[mogley76]

Homework Statement

a ray of light traveling through glass is incident at the interface with air at an angle of 30 deg to the normal. if the ray is refracted at the critical angle, what is the refractive index of the glass?

Homework Equations

none

The Attempt at a Solution

sin (90)= n2/n1 therefore n = 1 is that right?

 

[frozenguy]

n of the glass can't equal 1, because that is for vacuum. And air is like n=1.0003 or something .

If the light incident on a barrier is incident at an angle less than the critical angle, light will refract through. If its more than the critical angle, the light will internally reflect.

This holds for light incident on an interface traveling in a material with a greater index of refraction than the material the light wants to enter.

So for this to even be possible, you know n of glass is going to be bigger than 1.0003.

Do you know snells law? You need that equation and the equation for the critical angle.
Do you have a reference to find those and read what comes with it?

 

[mogley76]

sin crit angle =n2 right?

 

[frozenguy]

sin crit angle =n2 right?

Not quite.

$$sin\left(\theta_{c}\right)=\frac{n_{2}}{n_{1}}$$

So $$\theta_{c} = sin^{-1}\left(\frac{n_{2}}{n_{1}}\right)$$
n₁ is the medium the light is in, n₂ is the index of air which is 1.0003

 

[rl.bhat]

sin crit angle =n2 right?

You can use a general expression
n₁sin(θ₁) = n₂sin(θ₂)
suffix 1for the incident medium and 2 for the refracted medium.
When the angle of incidence in the denser medium is equal to the critical angle, the angle of refraction in the rarer medium is 90^o.

 

[SammyS]

The Attempt at a Solution

sin (90)= n2/n1 therefore n = 1 is that right?

This should be: sin(30°)/sin(90°)= n₂/n₁ , where n₂ is the refractive index of air.