Question

In: Physics

A light shines from the bottom of a pool of water that is 65.5 cm deep....

A light shines from the bottom of a pool of water that is 65.5 cm deep.

How far away, relative to the spot directly above it, must the beam of light strike the air-water interface in order that the light does not exit the water? cm

Expert Solution

since the substance is water .. i assume the substance above the water is air ..

we have then n1 = index of water = 1.333.. n2 = index of air = 1

A = arc sin(n2/n1) = arc sin(1/1.333)

in case you dont know arc sin is a way of writing inverse sine function... it means to find the angle given the sine of the function

we are then from this equation told that the sine of the function is 1/1.333

we also know

sin A = opposite/ hyp

since we only are given ... the distance to the water surface its easier to use tan of the angle so easier to find the A then calculate the tangent then calculate the distance

A = arc sin(0.75018)
A = 0.84833 radians

tan 0.84833 = 1.13451

tan A = opposite / adjacent

we need the opposite side length ... because the angle is the angle from the normal or shortest line to the surface..

solve for opposite

opposite = adjacent * tan A = 65.5* 1.134515 = 73.31078 cm

so the light must be aimed 73.31 cm from straight to the surface

genius_generous answered 4 months ago

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