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# light in a tank of water Watch

1. A point source of light is at the bottom of a 30 cm deep tank of water. A black
disk of diameter D is floating on the water’s surface with its centre directly above
the light such that none of the light from the source leaves the water. Calculate
the minimum diameter of the disk.
The refractive index of water is nw = 1.33.
3. (Original post by samueltaylor123)
A point source of light is at the bottom of a 30 cm deep tank of water. A black
disk of diameter D is floating on the water’s surface with its centre directly above
the light such that none of the light from the source leaves the water. Calculate
the minimum diameter of the disk.
The refractive index of water is nw = 1.33.
Here's a hint. First draw a diagram.

If no light is visible above the surface, what happens to the ray shown? (And all other rays with an angle greater than theta.)
So what is the significance of the angle theta?
There is a formula for this special angle in terms of the refractive index of the water.
Once you have it, the geometry of the triangle will give you the length (D/2)

4. (Original post by Lord Nice)

5. critical angle!
6. yes i see how critical angle formula is involved, but what about rays greater than theta which bounce of the tank walls and hit the water air boundary. their angle of incidence wouldnt neccessarily be the critical angle would it?
7. (Original post by samueltaylor123)
yes i see how critical angle formula is involved, but what about rays greater than theta which bounce of the tank walls and hit the water air boundary. their angle of incidence wouldnt neccessarily be the critical angle would it?
The question doesn't mention these and I am certain you do not need to consider them. As it says "no rays leave the tank", you can assume that these other rays have been conveniently ignored.

On further reflection about this problem, I think you'll find that if you look at the geometry of reflected light from the sides of the tank, it is actually impossible for these rays to leave the surface, because their angle, after reflection, will always be greater than the critical angle.
8. thank you

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