How to Calculate the Horizontal Distance of Light in Water Using Snell's Law?

[elitespart]

1. Depth of a lake is 1637m. If a beam of light with an angle of incidence of 60.0º enters the water from the air, what is the horizontal distance between the point where the light enters the water and the point where it strikes the lake's bottom?


2. Snell's Law: ni(sin Өi) = nr(sin Өr)



3. I'm really lost on this one. I know the angle of incidence and index of refracion is 1(?) How do I go about getting horizontal distance. Is there some equation that I'm not considering? Any help would be appreciated.

 

[Werg22]

Can you calculate the angle of refraction?

 

[mjsd]

you need to work out \theta_r from the n_i, n_r ,\theta_i, then use trig to work out horizontal displacement

 

[elitespart]

Can you calculate the angle of refraction?

Yeah, it's 40.5º.OK that helps. But I forgot the trig part of it. Can someone jog my memory? Is it tan(40.5 x 1637)

 

[Werg22]

You can construct a triangle with the angle of refraction, the height of the lake and the horizontal distance now, can't you?

 

[elitespart]

Yeah man I can. So, I'm getting 1.63 m for an answer. Right...wrong?

 

[Werg22]

Hummm wrong. Look. The refraction angle is measured from the normal line. The normal goes from the surface of lake to the bottom. The length is 1637. Now look at this triangle......../|
....../..|
......../__.|

See the right side? This is the height. The bottom side is the horizontal distance and the angle between the left and the right side is the refraction angle. Now you use trigonometry to get the horizontal distance.

 

[.ultimate]

tan 40.5 = x/depth of lake ( perpendicular by base)

now calculate x by putting value of tan40.5

 

[elitespart]

Yeah that's what I did. I'm probably just screwing up in the math somewhere. So the angle of refraction is 40.5. The horizontal is opposite to it and the adjacent is 1637 m. So it'd be tan (40.5) = x / 1637. Now don't I just solve for x?

 

[.ultimate]

Answer = 1398.13 m

They are askin for distance between point of incidence and point of emergence (in ur case its bottom of the lake) . just draw a perpendicular at the incidence wrt the interface

 

[elitespart]

oo damn I got that. Thought it was wrong. Alrite thanks bro. Appreciate it.