Important SAT 2 material(like practice tests) free!

[Toxinwolf]

im gonna upload SAT 2 (chem,math,phy only) Imp Material... A large amount of practice tests are comming!
Dont forget to like and comment


sorry due to personal reasons, these are postponed....
Check back in a month...(Feb)
Really very sorry.

 

[unique111]

sagar65265 bro! I need your help here. The answer is also posted, but i didn't get it.

 

[Zuhaib Saadat]

umm dude, where are the materials? ._.

 

[unique111]

sagar65265 bro! I need your help here. The answer is also posted, but i didn't get it.

sagar65265 bro, help me here?

 

[sagar65265]

sagar65265 bro, help me here?

Right, then. When light passes from one medium to another, the rate at which the light travels through the new medium (i.e. it's speed in that new medium) changes, depending on certain properties of that medium. So, that property is given a value we commonly call it's "refractive index", and it is denoted simply by n.

So, when a ray of light travels from one medium to another, the ratio of speeds in each one of the media is given by the formula

n(2)/n(1) = v(1)/v(2)

and can be said to be a statement of Snell's law (not sure if you need to know this; the formula is common, the name may not be).

In the case of light travelling from a vacuum, or ordinary, transparent air, to any other material, we can simplify the equation by using the fact that the refractive index n for air or vacuum is equal to 1.0, and the velocity of light in air or vacuum is approximated by c = 3 * 10^8 meters per second. So, we can write

1 * c = n(2)v(2)
v(2) = c/n(2)

Writing n(2) as just n,

v(2) = c/n

Which is the formula they have also mentioned. Now, using the formula v = f λ , we can see that for the velocity of the wave to change, either one of the wavelength or the frequency of the wave must change from it's initial value; another fact is that the color of light we perceive depends on the frequency of that light (in basic quantum mechanics, the formula E = hf is mentioned regularly, and it is because photons have an energy predominantly determined by the frequency of their oscillation that we can differentiate between colors).

Suppose we embedded a camera in a piece of glass, it would be absurd for it to see the sun as green or leaves as purple; so, we can guess from there that the frequency of a wave does not change when it is refracted.

So, since frequency remains constant, we can replace v(2) and c in the equation above with f λ(2) and f λ(1) respectively to get

f λ(2) = f λ(1)/n

Cancelling the f,

λ(2) = λ(1)/n

So, since λ(1) = 6 * 10^7 m (it isn't obvious that this is so; maybe there might be a table of values for the wavelengths of different colors of light, but if there isn't I can't see how they'd expect anyone to solve this, unless they expect you to memorize all those values!)

λ(2) = (6 * 10^7)/(1.5) = 4 * 10^7 meters

Hope this helped!
Good luck for all your exams!

 

[unique111]

Thank you! Yeah, i got confused as the answer suggests you to use the value 6*10^7, and thought maybe you have to conduct a separate set of calculations to first obtain that value of λ.
Thanks for your help! may i ask you which level you're studying in?

 

[sagar65265]

Thank you! Yeah, i got confused as the answer suggests you to use the value 6*10^7, and thought maybe you have to conduct a separate set of calculations to first obtain that value of λ.
Thanks for your help! may i ask you which level you're studying in?

You're welcome, though I should say this - if there is some sort of preliminary calculation to obtain the wavelength of orange light, I have no idea what it is, so the best way of confirming would be checking the entire paper for a table containing those values or an additional information sheet for these values. Supposing it's there, then there should not pose any problems .

I've finished my A levels, btw