Optics-Critical Angle in Different Mediums

[physgrl]

Homework Statement

11. The critical angle of a piece of transparent material in air is 37.30. What is the critical angle of the same material when it is immersed in water? (n = 1.33)

a. 41.40
b. 48.40
*c. 53.70
d. 63.00
e. 68.20

Homework Equations

βc=sin-1(n2/n1)

The Attempt at a Solution

n2=n1*sinβc-in-air
n2=sin(37.3)

βc-in-water=sin-1(n2/n1)
βc-in-water=sin-1(sin(37.3)/1.33)
βc-in-water=27.1°

what am I doing wrong they say the answer is 53.7° thanks!

 

[gneill]

Remember that the critical angle is the angle at the threshold of complete internal reflection. Thus for the initial scenario it is the angle measured inside the block of material, not in the air.

Complete internal reflection can only occur when the index of refraction on the far side of the boundary is less than the index of refraction in which the light is currently traveling.

 

[physgrl]

ohh so n2 is really n1 basically

 

[physgrl]

i get an error :/

 

[gneill]

i get an error :/

You'll have to show your work so we can see.

 

[physgrl]

n1=n2*sinβc-in-air
n1=1/sin(37.3)

βc-in-water=sin-1(n2/n1)
βc-in-water=sin-1(1/sin(37.3)*1.33)
βc-in-water=error

 

[gneill]

n1=n2*sinβc-in-air
n1=1/sin(37.3)

βc-in-water=sin-1(n2/n1)

Good so far.

βc-in-water=sin-1(1/sin(37.3)*1.33)

Oops. I think the 1/sin bit has thrown you for your n2/n1 expression. n2 here is that of water, 1.33, so it should be in the numerator. 1/sin(37.3) should comprise the denominator.

Why not just go with a numerical value of n1 from the previous step?

 

[physgrl]

so then it should be:
βc-in-water=sin-1(sin(37.3)*1.33)

so n2 is the material from which it comes from if β2 is the critical angle and in this case i comes from the water/air to the mysterious medium right? i think i was confusing which medium was 1 and which was 2 in my mind

 

[gneill]

so then it should be:
βc-in-water=sin-1(sin(37.3)*1.33)

so n2 is the material from which it comes from if β2 is the critical angle and in this case i comes from the water/air to the mysterious medium right? i think i was confusing which medium was 1 and which was 2 in my mind

In both cases the light is moving from within the mystery medium towards the interface with the air or water. Check the numerical value that you got for the index of refraction of the mystery medium. I think you'll find it to be larger than that of both air and water.

 

[physgrl]

n=1.65

so its basically critical=sin-1(n-to/n-from) right? cause the critical is like angle-from and the 90o is the angle-to

 

[gneill]

That's right.

I usually start with the general Snell's law equation: n1*sin(θ1) = n2*sin(θ2) and remember that for a critical angle to occur the light must be going from the medium with the larger index of refraction to the one with smaller index of refraction. Let the starting medium be "n1". Then plug in the conditions for critical angle:

n1*sin(θ_crit) = n2*sin(90°)

n1*sin(θ_crit) = n2

Then solve for whatever it is you're looking for.

 

[physgrl]

makes sense! Thanks! :)