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# What is this supposed to mean?

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1. http://www.bbc.co.uk/schools/gcsebit...es/revision/2/

Step 1: Work out the sine of angle isin 55 = 0.819Step 2: Work out the sine of angle rsin 33 = 0.545

How does sin automatically = that? Bare with me I am TERRIBLE AT MATHS. My brain can't seem to understand it. Good at English/Art though so I must be a right sider.

And also;;http://www.bbc.co.uk/bitesize/higher...on/revision/2/

QuestionNow calculate the angle of refraction of the ray.We know thatAndHideAnswerSoTherefore, Angle of refraction = 35°

How the hell am I supposed to know what sin 0 is from the previous info?

THANKS!
2. Ok I figured out a bit

but here

at 3;30 he says plug that into your calculator,,,,, I have no Idea how to do that?
3. (Original post by DNBK)
Ok I figured out a bit

but here

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?
4. (Original post by Dilzo999)
Inverse sine on your calculator. You should also be learning how to do these things in your maths class. Are you paying attention?
It's Physics. How do you inverse sin? Ie what's the button look like?
5. Ah figured it out. It's really hard for me.
6. RIGHT NOW I HAVE A BIGGER PROBLEM. How do you calculate angle of incidence? Is it just like refraction but reversed?
7. (Original post by DNBK)
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:

8. (Original post by uberteknik)
Yes. Angle of incidence = angle of reflection. Be careful where the angles are measured:

Thanks!

also this:

Now calculate the wavelength of the light in air

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?
9. (Original post by DNBK)
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?

10. (Original post by The-Spartan)

I did but still to the power of twenty one?
11. If your getting to the power of 21 your putting x10-14 on the bottom. You should roughly get x10-6 not got a calculator to hand
12. ah figured it. A lot of this is not being taught how to use a scientific calculator much, I put the 5.40 part in brackets then divided by the 1.50

SORRY GUYS MY PHYSICS TEACHER IS NOT VERY GOOD!
13. The 1uestion I have now is, when to use brackets? All the time?
14. 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°
15. (Original post by DNBK)
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 180o.

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 70o:

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

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

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

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

Then from the diagram above:

The unknown value here is

Rearrange the expression to get on its own:

and plugging the known values in

finally

This answer (40o) is the angle of refraction as the light (entering the prism at an incidence angle of 70o) 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 180o, straight line = 180o, right angle = 90o etc).

Just follow it through and be very careful about the definitions above namely the angles and and their relationship in the expression:

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.
16. (Original post by uberteknik)
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 180o.

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 70o:

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

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

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

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

Then from the diagram above:

The unknown value here is

Rearrange the expression to get on its own:

and plugging the known values in

finally

This answer (40o) is the angle of refraction as the light (entering the prism at an incidence angle of 70o) 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 180o, straight line = 180o, right angle = 90o etc).

Just follow it through and be very careful about the definitions above namely the angles and and their relationship in the expression:

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!
17. (Original post by DNBK)
THANK YOU! for your very detailed response!
How did you get on with finishing the question?

Once you know how to split up the problem, all you really need is:

together with working out the incident angle of the light arriving at the boundary between the plastic of the prism and the exit vacuum.

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