Question

Imagine you are 2.8 m below the surface of a pool of water looking up toward the surface. How far in front of you would you need to look before you can no longer see through the surface of the water, i.e. the surface starts to appear mirror-like? Assume there is air above the surface.

How far from a converging lens with a focal length of 41 cm should an object be to produce a real image that is 2.5 times the size of the object?

Answer 1

1)
The critical angle for total internal reflection of water-air interface is 48.6 degrees, This can be worked out as follows,
Using Snell's law,
n1 * sinc = n * sinr
Where n1 is the refractive index of water, c is the critical angle, n is the refractive index of air and r is the refracted angle.
For total internal reflection, r = 90 degrees
Substituting values,
1.333 * sinc = 1 * sin(90)
c = sin^-1(1/1.333)
= 48.6 degrees
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tanc = x / h
Where x is the maximum horizontal distance we can see through the surface and h is the height below water surface.
x = h * tanc
= 2.8 * tan(48.6)
= 3.2 m

2)
Consider the object distance as u and the image distance as v. For a real image, the object must not lies within the focal length of the lens.
Given that magnification is 2.5
v / u = 2.5
v = u * 2.5

1 / f = 1 / u + 1 / v
Where f is the focal length.
Substituting values,
1 / 41 = 1 / u + 1 / (2.5 * u)
1 / 41 = 1 / u * (1 * 1 / 2.5)
u = 41 * 1.4
= 57.4 cm