Imagine you are 2.8 m below th
Imagine you are 2.8 m below the surface of a pool of waterlooking up toward the surface. How far in front of you would youneed to look before you can no longer see through the surface ofthe water, i.e. the surface starts to appear mirror-like? Assumethere is air above the surface.
How far from a converging lens with a focal length of 41 cmshould an object be to produce a real image that is 2.5 times thesize of the object?
Answer:
1)The critical angle for total internal reflection of water-airinterface is 48.6 degrees, This can be worked out as follows,Using Snell’s law,n1 * sin
c= n * sinrWhere n1 is the refractive index of water, c is the critical angle, n is the refractive index of air andr is the refracted angle.For total internal reflection, r = 90 degreesSubstituting values,1.333 * sinc= 1 * sin(90)c = sin-1(1/1.333)= 48.6 degrees———————————————–tanc= x / hWhere x is the maximum horizontal distance we can see through thesurface 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. Fora real image, the object must not lies within the focal length ofthe lens.Given that magnification is 2.5v / u = 2.5v = u * 2.5
1 / f = 1 / u + 1 / vWhere 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
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