What is this supposed to mean?

Ok I figured out a bit

but here
https://www.youtube.com/watch?v=vWzph654fDU

at 3;30 he says plug that into your calculator,,,,, I have no Idea how to do that?

Inverse sine on your calculator. You should also be learning how to do these things in your maths class. Are you paying attention?

RIGHT NOW I HAVE A BIGGER PROBLEM. How do you calculate angle of incidence? Is it just like refraction but reversed?

Yes. Angle of incidence = angle of reflection. Be careful where the angles are measured:

Answerv = c = 3.00 × 10⁸ ms¹
f = 4.8 × 10¹⁴ Hz
3.00 × 10⁸ = 4.8 × 10¹⁴
× = 6.25 × 10^-7 m.
I keep getting 6.25x10 to the power of 21 not minus seven? Am I entering it into the claculator correctly?

$c=f\lambda$

$\frac{c}{f}=\lambda$

$\dfrac{3.00 \times 10^8}{4.8 \times 10^{14}}=\lambda$

Type this in, you should get your answer.

NOW I HAVE NO IDEA HOW TO WORK THIS OPUT???

A ray of red light is incident on a triangular prism made from clear plastic (as shown in the diagram above). The refractive index of the plastic is 1.46 for the red light. Calculate the angle of refraction as the ray of red light leaves the prism.23°30°35°

You can do it. It's much easier than it first looks.

Split (pun intended) the problem up by following the ray of light as it enters the prism and then leaves it.

But first, notice that the prism is equilateral (all internal angles and length of sides are equal). Also remember that the a straight line is effectively 180^o.

You will need to understand what the refractive index for the plastic means.

Refraction is different to reflection. The key thing is that unlike reflection, the incidence angle and refraction angle are not the same.

Refractive index diagram:



You also need to know what the relationships for angles, wavelength and velocity are which determine the refractive index for any material:



n₁ is the refractive index for a vacuum and you can take this to be 1.


FIRST PART: Refraction of red light entering the plastic prism at an incidence angle of 70^o:

So plug the values into the first line and you get:

$\frac{n_1}{n_2} = \frac{1}{1.46} = 0.685$

Now look at the incidence angle which the (question) diagram says is 70^o.



So you need to work out the refraction angle as the light enters the prism.

Notice the dotted normal line which is 90^o to the surface of the prism?

Then from the diagram above:

$\frac{n_1}{n_2} = \frac{sin\theta _2}{sin\theta _1} = \frac{1}{1.46} = 0.685$

The unknown value here is $sin\theta _2$

Rearrange the expression to get $sin\theta _2$ on its own:

$\frac{n_1}{n_2} = \frac{sin\theta _2}{sin\theta _1} = 0.685$

$sin\theta _2 = 0.685sin\theta _1$

and plugging the known values in

$sin\theta _2 = 0.685sin(70) = 0.685 \rm \ x \ 0.94 = 0.644$

$sin\theta _2 = 0.644$

finally

$\theta _2 = sin^{-1}( 0.644)= 40^o$


This answer (40^o) is the angle of refraction as the light (entering the prism at an incidence angle of 70^o) transitions from the vacuum to the plastic prism. You now need to follow the ray of light through the plastic as it exits the prism.

SECOND PART: Refraction of red light exiting the plastic prism:

Note you need to first find the incidence angle of that ray of light passing through the plastic and out into the vacuum.

All the information you need is given. Remember to use your knowledge about angles (internal angles of a triangle sum to 180^o, straight line = 180^o, right angle = 90^o etc).

Just follow it through and be very careful about the definitions above namely the angles $\theta _ 2$ and $\theta _1$ and their relationship in the expression:

$\frac{n_1}{n_2} = \frac{sin\theta _2}{sin\theta _1}$

i.e. the light now transitions from the prism to a vacuum so use basic angles to work out the angle of incidence to the exit boundary of the light from the plastic to vacuum.

See how you get on.

THANK YOU! for your very detailed response!