Images can be formed by reflection and by refraction. Assume that we have a boundary
between two media with different indices of refraction, and that the boundary
has a radius of curvature R, as shown in the figure. The center of
curvature is at the origin.
Light is refracted at the boundary. Two rays starting at an object are shown. One ray is incident normally and is not refracted. The other ray is incident at an angle θi with respect to the normal, and the refracted angle θt is given by Snell's law, nisinθi = ntsinθt.
An image is formed where the two rays intersect.
For small angles θ, (θ(rad) << 1) we have sinθ
≈ tanθ ≈ θ.
(See the graph on the right which plots θ(rad), sinθ, and tanθ versus θ(deg) for angles from 0 to 10o.)
This is called the small angle approximation. It simplifies calculations if it is applicable.
small angle or paraxial approximation we can show that for a spherical interface
n1/xo + n2/xi = (n2 - n1)/R,
if we adopt the following sign conventions.
The magnification is M = hi/ho = -(n1xi)/(n2xo). If M is negative, the image is inverted.
A simple model of the human eye ignores its lens entirely. Much of what
the eye does to light happens at the transparent cornea. Assume that this
outer surface has a radius of curvature of 6 mm, and assume that the eyeball
contains just one fluid of index of refraction n = 1.4. Prove that every
distant object will be imaged on the retina, 21 mm behind the cornea. Describe the image.
For distant objects xo is very large and n1/xo is very small.
We can safely ignore this term in the equation n1/xo + n2/xi = (n2 - n1)/R.
We then have n2/xi = (n2 - n1)/R,
which yields xi = n2R/(n2 - n1) = (1.4*6 mm)/0.4 = 21 mm.
The image distance is 21 mm.
The image is very small and inverted.
flat interface we have R = infinite and n1/xo + n2/xi = 0.
Let an object be underwater, 10 cm from the water-air interface.
We find xi = -(n2/n1)xo = -(1/1.33)xo. xi = -7.5 cm.
The magnification is M = 1, it is positive.
The image is a virtual, non-inverted image.
A detector viewing the object from the air side sees the image at the apparent position closer to the interface.
A cubical block of ice, 50 cm on a side, is placed on a level floor over
a speck of dust. Find the location of the image of the dust if the index of
refraction of the ice is n = 1.309.
xo = 50 cm. xi = -(1/1.309)xo = -38.2 cm.
The apparent position is 38.2 cm below the top surface of the ice.