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physics inquiry!!!

HorsePower

HorsePower

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ullahabd said:
Oliveme said:
the lines coming from the boundary would slant towards it. check the mark scheme. they have drawn it.
http://www.xtremepapers.me/CIE/Cambridg ... _ms_31.pdf
good luck :)
Click to expand...

but isnt there a particular method for this?
bcuz i recall a friend talking bout it
sorry i suck at physics.
Click to expand...
yes there is a particular method, but u dont have to do it in this question cuz the space they have given is really small for all the construction . plus the marks for the arent even that much, so just draw normally :)
 
Oliveme

Oliveme

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yeah do you mean the incident angle should be bigger than the refracted one? yes you do that when you have a bigger diagram. this is because the speed is decreasing as it goes into shallow water as the wavelength decreases. always remember, when waves go from deep water to shallow water, their speed decreases. :) hope i helped.
 
U

unplugged

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The way i always imagine it is that the wave is like half a circle. And when the wave is partially inside a slower medium and partially outside , you can imagine your finger pressing on the slower side but the wave keeps moving. so what happens is that it rotates and that is ALWAYS the right direction. So just rotate the line a bit in that direction and make the distance between the lines less because the medium is slower and make sure all the waves inside of it are parallel.
 
C

ChrisChin

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Wavefronts swing towards the normal if its speed is slower in the new medium, and away from the normal if its speed is faster in the new medium. Make sure you read the question carefully. If refractive index, critical angle, object/image lengths or speed of wave in the medium is given, use it to calculate the correct angle in respective mediums, using

n = sin i / sin r = 1 / sin x = speed1 / speed2 = object length / image length ; where

n = refractive index
i & r = wave angles; i > r
x = critical angle
speed1 > speed2
 
I

igcse2011

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i Need help with :
1) use and describe the use of a cathode-ray oscilloscope to display waveforms.
2) Describe experiments to show good and bad emitters and good and bad absorbers of infra-red radiation.
3) Explain why energy losses in cables are lower when the voltage is high.

Thanks in advance :)
 
C

ChrisChin

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3)

Power = IV = I^2xR

For transmission of any fixed amount of energy, energy transmitted per second = IV. By transmitting at 10x voltage, current is reduced to 1/10 or 10%, Power loss due to resistance in conductor = I^2R , => reduction of current to 1/10 will produce a reduction in power loss to 1/100 or 1% of original power loss.
 
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