[KurzickMushroom]

Hey guys, I got this optics problem I don't understand. I drew out the diagram but still don't understand which forumla to apply. I am extremely confused over why the index of refraction for water is given, I'm not sure where that comes in.   

A person stands beside his swimming pool, which is full of water. His eyes are at a level 1.6 m directly above the edge of the pool, and looks directly across the pool which is 7 m wide. Light from the bottom corner opposite him arrives at his eye in such a direction that the apparent depth of water, h, is 0.3 m. If the Refractive Index of water is 4/3, calculate H, the actual depth of the pool.


Thank you in advance.
-Kurzick Mushroom

[Yggdrasil]

Remember that light will refract or bend at the air-water interface and the angle that it refracts is dependent on the difference index of refraction between water and air.  You can calculate this angle using Snell's law.

[KurzickMushroom]

Remember that light will refract or bend at the air-water interface and the angle that it refracts is dependent on the difference index of refraction between water and air.  You can calculate this angle using Snell's law.

They don't give you any angles. You need at least one angle describing the ray before you can apply Snell's law. Is it implied or stated in the question?

[zxt]

sinA/sinB=4/3,A is the incident angle and B is refractional angel. Also, you should find out the incident point where the normal stands. Next, there are similar triangles for you to calculate the angles and lengths and finally my answer is 1.06m, right or not?

[KurzickMushroom]

sinA/sinB=4/3,A is the incident angle and B is refractional angel. Also, you should find out the incident point where the normal stands. Next, there are similar triangles for you to calculate the angles and lengths and finally my answer is 1.06m, right or not?

Your answer is right, thank you but I'm still trying to understand this...

[Yggdrasil]

They don't give you any angles. You need at least one angle describing the ray before you can apply Snell's law. Is it implied or stated in the question?

You should be able to calculate the angle from the values given (using trigonometry).  When you draw your diagram, remember that your eye will assume that light from the apparent bottom of the pool had traveled in a straight line through the air-water interface, whereas the light had really bent at the air-water interface.

[zxt]

Find out 7-x and tanC. Hope this helps.

[mogley]

i still cant get that answer...how do you get that answer??