Question

Please Help!

A ray of violet enters a 60 degree glass prism at an angle of incidence of 40 degrees. If the refractive index of the prism material for violet light is 1.651, calculate

a) angle of refraction for the air-glass boundary

b) the critical angle for the violet light leaving the glass

c) the velocity of the violet light in the prism

Even some hints would be appreciated, I would love to see the working though thanks.

Answer 1

for a) I got 1.651 = sin(60degrees) / sin(r)
so angle of refraction = 31.6 degrees

Answer 2

What you need: Snell's law
\[\frac{Sin(\Theta_{i})}{Sin(\Theta_{r})}=\frac{n_{r}}{n_{i}}\] The subscripts indicate the whether it's incident or refracted.

a) Air to glass boundary, so
\[n_{r} = 1 (Air), \Theta_{r} = 40^{o}\]\[n_{i} = 1.651 , \Theta_{i} = ?\] Subbing these values and solving for Theta i, the answer is 22.9° from the normal.

b) Critical angle only exists when light enters a medium of higher refractive index. This is because as light enters a medium of higher refractive index, light bends away from the normal, and the maximum allowed bending is 90 ° from the normal. Beyond the crtical angle (of incidence) it will obey the law of reflection instead. Thus I will use snell's law and rearrange the terms\[\frac{Sin(\Theta_{c})}{Sin(90^{o})}=\frac{1}{1.651}\]Solving, the answer is: 37.3°

c) The refractive index, n, is simple the ratio of the speed of light in vacuum to the speed of light in another medium. There is an underlying assumption that the speed of light in vacuum can never be violated, and it is currently accepted that you cannot break this cosmic speed limit. You can be faster than the speed of light in glass, but not faster than the speed of light in vacuum. The relation is thus:
\[\frac{c}{v}=n\] Where c is the speed of light in vacuum and v is the speed of light in a medium. Since v cannot be greater than c, that means n must always be larger than 1.
Putting in the values for c=3*10^8, and n = 1.651, v= 1.82*10^8m/s

Answer 3

Yeah, (a) is correct.
For (b), do you know what is critical angle?

Answer 4

hey guys thanks.... I just started learning this topic so not really too confident critical angle is the maximum angle that light will not internally reflect right?

Answer 5

Sin i(c)= Sin(90 degrees)/1/1.651

so i(c)= 37.3 degrees

Answer 6

Hi @Festinger thanks for your reply, the way you did this looks different from what I was taught so I'm still studying it. for example the index 1.651 I thought was already the ratio between Air and glass, therefore I thought I didn't need the value for air? I'm confused :S

Answer 7

Yes, critical angle is the maximum angle from the normal of the plane which the light is incident on.

You are correct to say that you don't need it between air and glass, but if the question is between water and glass, and you are only given the refractive index between air and glass and air and water, then you still can use the more general form of solving I have shown to solve the problem!

As a hypothetical example, I let the refractive index be 1.53 be glass, and 1.33 be water. When light passes from glass to water, I water is easier for light to pass, so it "bends away" from the incident path.

Going back to the form I gave (which is snell's law):
\[\frac{Sin(\Theta_{c})}{Sin{90^{o}}}=\frac{n_{r}}{n_{i}}\]

Sin 90 is 1, and the index of the refracted beam is now 1.33 (remember, we are going from glass to water!) and 1.53 for the incident beam.

So, putting in the relevant info:
\[\frac{Sin(\Theta_{c})}{1}=\frac{1.33}{1.53}\]

Which is about 60.4 degrees!
We can define a refractive index between water and glass, but it won't be that meaningful as we can calculate it if we know the refractive index of the respective material in air (which is taken to be 1, though it's something like 1.0003)

Answer 8

I found a java applet which you can play with: http://interactagram.com/physics/optics/refraction/

To change the angle of incidence simply click on the touch and drag it up and down, and click on the button to change the material, and hence refractive index.

Also, what I wanted to show you is also how we obtain the formula for critical angle, which is basically snell's law. I hated remembering formulas, and thus I prefer to remember a general formula, and work my way from there.

Answer 9

THANK YOUS!!