Help on physics problem

[Jess]

There's just one physics problem I am completely STUMPED on. I need to get help by tomorrow, before physics class. so please post asap.

If you don't know the answer then don't post please!

the physics problem is based on law of refraction, if that's any help

Here's the problem:

In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight 1.3m above the water level, onto the surface of the water at a point 2.7m from his foot at the edge of the pool. Where does the spot of light hit the bottom of the pool, relative to the edge, if the pool is 2.1m deep?

[Mortal Coil]

In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight 1.3m above the water level, onto the surface of the water at a point 2.7m from his foot at the edge of the pool. Where does the spot of light hit the bottom of the pool, relative to the edge, if the pool is 2.1m deep?

So, he is 1.3m high, shining his flashlight 2.7m out. This means that the angle of incidence is 90-(tan theta= 1.3/2.7) = 64.3 degrees.

The refractive index of water is 1.3330, meaning that sin(64.3)/sin(angle of refraction) = 1.3330. sin(64.3)/1.3330=0.6759...
sin^-1(ans)=42.53

Now we have a right angle triangle where the smallest angle is 42.53 degrees and one side is 2.1m. In order to find the length along the bottom, you have to use trig again. So, tan(42.53)=x/2.1
tan(42.53)*2.1=1.92631......
Therefore, the distance from the edge is 2.7+1.9
2.7+1.9=4.6 meters.

PS. Jess, I'm ashamed of you. You're Chinese :P

[Peace God]

please tell me your teacher wants you to use snell's law of refraction, if not ignore this :P

http://img707.imageshack.us/img707/8035/b97a1955f2694b7e867feef.jpg

Edit: DSKLJBGASKLBJASGDASDFBGJUASLDBG BEEAT MEE TO IT!!!...haha its cool tho

[Jess]

managed to figure it out like 5 minutes after I posted this. pfft. but thank you both anyways :p

So, he is 1.3m high, shining his flashlight 2.7m out. This means that the angle of incidence is 90-(tan theta= 1.3/2.7) = 64.3 degrees.

The refractive index of water is 1.3330, meaning that sin(64.3)/sin(angle of refraction) = 1.3330. sin(64.3)/1.3330=0.6759...
sin^-1(ans)=42.53

Now we have a right angle triangle where the smallest angle is 42.53 degrees and one side is 2.1m. In order to find the length along the bottom, you have to use trig again. So, tan(42.53)=x/2.1
tan(42.53)*2.1=1.92631......
Therefore, the distance from the edge is 2.7+1.9
2.7+1.9=4.6 meters.

PS. Jess, I'm ashamed of you. You're Chinese :P

thanks. and lol just because I'm Chinese doesn't mean I have to be super smart :D :rolleyes:

please tell me your teacher wants you to use snell's law of refraction, if not ignore this :P

image (http://img707.imageshack.us/img707/8035/b97a1955f2694b7e867feef.jpg)

Edit: DSKLJBGASKLBJASGDASDFBGJUASLDBG BEEAT MEE TO IT!!!...haha its cool tho

thank you anyways. and yeah it involves Snell's Law

[Peace God]

right then glad that's settled. this thread is now about my groovy 70s city college of san fransisco textbook that i found. its ok to be jelly guys...

http://img707.imageshack.us/img707/857/97575221.jpg

http://img208.imageshack.us/img208/8450/30164232.jpg

http://img819.imageshack.us/img819/9705/45078264.jpg
^ lol refraction

dont judge me. i never been able to show it off since i got it :props: