In: Physics
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
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