In the physics laboratory, a narrow beam of light passes through a glass prism and makes a bright spot on the wall. Turn the prism slowly on the table, always the same way. The spot moves along the wall. Then it slows, stops, and moves back the way it came, although you are still turning the prism the same way.
Two refractions, both towards the base
A prism is a block of glass with two flat faces inclined to each other. The angle between them is the angle of the prism \(A\), and the side opposite that angle is the base. A ray meets the first face at the angle of incidence \(i\) and refracts to \(r_1\). It crosses the prism, meets the second face at \(r_2\), and leaves at the angle of emergence \(e\).
At each face the ray goes from one medium to the other and bends. At the first face it enters the glass and bends towards the normal. At the second it leaves the glass and bends away from the normal. Both bends turn the ray the same way, towards the base. The total turn is the angle of deviation \(\delta\).
\[ \text{angle between the two normals} = 180° - A \]
Why. each normal is at \(90°\) to its face. In the quadrilateral formed by the two faces and the two normals, the angles add to \(360°\): \(A + 90° + 90° + (\text{angle between normals}) = 360°\).
\[ r_1 + r_2 + (180° - A) = 180° \]
Why. the ray inside and the two normals form a triangle, whose angles at the faces are \(r_1\) and \(r_2\).
\[ r_1 + r_2 = A \]
Why. rearrange step 2.
\[ \delta = (i - r_1) + (e - r_2) \]
Why. the first face turns the ray by \(i - r_1\), the second by \(e - r_2\), both towards the base, so the turns add.
\[ \delta = i + e - (r_1 + r_2) = i + e - A \]
Read it. substitute step 3.
A ray falls at \(i = 60°\) on one face of a prism with \(A = 60°\) and \(\mu = \sqrt3\). Find \(r_1\), \(r_2\), \(e\) and the deviation. A second ray enters a \(30°\) prism of \(\mu = \sqrt2\) normally. Find its angle of emergence and deviation.
First prism: \(\sin r_1 = \sin 60°/\sqrt3 = \tfrac12\), so \(r_1 = 30°\) and \(r_2 = 60° - 30° = 30°\). Then \(\sin e = \sqrt3\sin 30° = \sqrt3/2\), so \(e = 60°\), and \(\delta = 60° + 60° - 60° = 60°\).
Second prism: normal incidence means \(i = 0\) and \(r_1 = 0\), so \(r_2 = 30°\). \(\sin e = \sqrt2\sin 30° = 1/\sqrt2\), so \(e = 45°\), and \(\delta = 0 + 45° - 30° = 15°\).
A ray enters a prism of angle \(60°\) at \(i = 45°\) and leaves at \(e = 52.4°\). What is its deviation, in degrees?
The deviation has a minimum
As the prism turns, the angle of incidence changes, and so does the deviation. Most students expect the deviation simply to grow, or simply to fall, as \(i\) grows. Predict it before you play the figure.
A ray passes through a glass prism (\(\mu = 1.5\), \(A = 60°\)). The angle of incidence grows from \(27.9°\) to \(90°\). How does the deviation change?
The curve in the figure above starts at \(i = 27.9°\) with \(\delta = 57.9°\), when the ray leaves grazing the second face. It falls to a minimum of \(37.2°\) at \(i = 48.6°\). There the ray inside runs parallel to the base, and \(e = i\). Then it rises again to \(57.9°\) at grazing incidence, \(i = 90°\). The two ends have the same deviation, joined by the dashed line.
Why is the minimum at the symmetric path? Reverse any ray. It retraces its path, with \(i\) and \(e\) swapped and the same deviation. So every deviation, except one, is produced by two angles of incidence, \(i\) and \(e\). The single lowest point of the curve cannot come in such a pair. It must be the one path that is its own reverse, with \(i = e\).
The deviation of a ray through a prism is least when the ray passes symmetrically through it. Then \(i = e\), \(r_1 = r_2 = A/2\), and the ray inside is parallel to the base for an isosceles prism. This least value is the minimum deviation \(\delta_m\).
At the minimum, \(r_1 = r_2 = A/2\) and \(i = e\). The relation \(\delta_m = 2i - A\) gives \(i = (A + \delta_m)/2\). Snell's law at the first face then gives the refractive index from two angles that are easy to measure.
For glass of \(1.5\) and \(A = 60°\), \(\sin i = 1.5\sin 30° = 0.75\), so \(i = 48.6°\) and \(\delta_m = 2 \times 48.6° - 60° = 37.2°\). For the round value \(\mu = \sqrt2\), \(\sin i = \sqrt2 \times \tfrac12\), so \(i = 45°\) and \(\delta_m = 30°\). Measuring \(A = 60°\) and \(\delta_m = 30°\) in the laboratory gives \(\mu = \sin 45°/\sin 30° = \sqrt2\).
For an equilateral prism, the minimum deviation is found to be equal to the angle of the prism. What is the refractive index of the glass?
Explain in two or three sentences, using the reversibility of light, why the deviation of a prism is least when \(i = e\).
If a ray entering at \(i\) leaves at \(e\), the reversed ray enters at \(e\) and leaves at \(i\) with the same deviation. So every deviation is produced by two angles of incidence, unless \(i = e\). The curve of \(\delta\) against \(i\) has one lowest point, which cannot be shared by two angles, so it must be at the symmetric path, \(i = e\).
Now return to the turning prism in the laboratory. As you turn it, the angle of incidence grows steadily. Past the symmetric position, the deviation stops falling and starts to grow. So the spot on the wall moves, stops at the minimum deviation, and comes back. Measure that turning point and the prism angle, and the formula gives \(\mu\).
Why the curve does not start at zero
The deviation curve began at \(27.9°\), not at \(i = 0\). At a small angle of incidence, \(r_1\) is small, so \(r_2 = A - r_1\) is large. If \(r_2\) is larger than the critical angle \(C\), the ray is totally reflected at the second face and never leaves. The limit is \(r_2 = C\). Then \(e = 90°\): the ray leaves grazing the face. This is grazing emergence.
For glass of \(1.5\), \(C = 41.8°\). With \(A = 60°\), grazing emergence needs \(r_1 = 60° - 41.8° = 18.2°\). Then \(\sin i = 1.5\sin 18.2° = 0.468\), so \(i = 27.9°\). No ray below \(27.9°\) gets out. For \(\mu = \sqrt2\) in the same prism, \(C = 45°\), \(r_1 = 15°\) and the least angle is \(21.5°\).
Can a prism trap every ray? At the first face \(r_1\) is at most \(C\), reached at grazing incidence. To get out, the ray also needs \(r_2 \le C\). So it needs \(A = r_1 + r_2 \le 2C\). A prism with \(A > 2C\) lets no ray through at all. For \(\mu = 2\), \(C = 30°\): a prism of \(60°\) passes only the grazing ray, which also leaves grazing, and anything steeper passes nothing. Even a \(30°\) prism of this material is at the limit for normal incidence: \(r_1 = 0\), so \(r_2 = 30° = C\), and the ray leaves grazing.
A ray falls at \(i = 45°\) on a \(60°\) prism of \(\mu = \sqrt3\). Take \(\sin^{-1}(0.41) \approx 24°\) and \(\sin^{-1}(1/\sqrt3) \approx 35°\). Does the ray emerge from the second face?
Find \(r_1\) from Snell's law, then \(r_2 = A - r_1\), and compare \(r_2\) with the critical angle.
\(\sin r_1 = \sin 45°/\sqrt3 = 1/\sqrt6 \approx 0.41\). \(\sin C = 1/\sqrt3\).
\(r_1 \approx 24°\), so \(r_2 \approx 36°\). \(C \approx 35°\). So \(r_2 > C\).
The ray is totally reflected at the second face and does not emerge. At \(i = 60°\), the same prism let the ray through with \(r_2 = 30°\), at minimum deviation.
A prism is made of a material with \(\mu = 2\). What is the largest prism angle, in degrees, that lets any ray through?
A ray falls at grazing incidence (\(i = 90°\)) on a prism with \(A = 90°\) and \(\mu = \sqrt2\). What happens at the second face?
The thin prism
Many prisms in instruments have a small angle, a few degrees. For such a thin prism, and for rays near normal incidence, all the angles are small. Then \(\sin\theta \approx \theta\) in radians, and the deviation becomes very simple.
\[ i = \mu r_1, \qquad e = \mu r_2 \]
Why. Snell's law at each face, with \(\sin\theta \approx \theta\) for small angles.
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Your turn. substitute step 1 into the deviation. Which line is it?
\[ \delta = \mu A - A = (\mu - 1)A \]
Read it. \(r_1 + r_2 = A\) for every prism.
The result does not contain \(i\). A thin prism deviates every ray near normal incidence by the same small angle. A glass prism of \(4°\) and \(\mu = 1.5\) turns light by \((1.5 - 1) \times 4° = 2°\). Two thin prisms in contact add their deviations when they point the same way and subtract them when one is turned round.
A thin prism of \(6°\) and \(\mu_1 = 1.5\) is placed in contact with another of \(4°\) and \(\mu_2 = 1.6\). Find the net deviation when the prisms point the same way, and when they point opposite ways.
\(\delta_1 = 0.5 \times 6° = 3°\) and \(\delta_2 = 0.6 \times 4° = 2.4°\).
Same way round, the deviations add: \(3° + 2.4° = 5.4°\).
Opposite ways round, they subtract: \(3° - 2.4° = 0.6°\), towards the base of the first prism.
A thin glass prism of \(8°\) and \(\mu = 1.5\) deviates light by \(4°\) in air. What is its deviation, in degrees, when it is under water (\(\mu = \tfrac43\))?
In the laboratory, the angle of a prism is measured with a spectrometer. A parallel beam falls on the apex, and each face reflects part of it like a plane mirror. The normals to the two faces are inclined at \(A\), so the incident beam makes angles \(\theta_1\) and \(\theta_2\) with them, where \(\theta_1 + \theta_2 = A\). A mirror turns a beam through twice its angle of incidence, so the two reflected beams are turned \(2\theta_1\) and \(2\theta_2\) to opposite sides. The angle between them is \(2A\): the telescope reads twice the prism angle.
- Writing \(\delta = i + e + A\), or forgetting to subtract \(A\).\(\delta = (i - r_1) + (e - r_2) = i + e - A\). The prism angle is subtracted once.
- Using \(\mu = \sin\frac{A + \delta}{2}/\sin\frac A2\) at any deviation.The formula needs the symmetric path, \(i = e\). It holds only at minimum deviation.
- Expecting the deviation to fall steadily as \(i\) grows.It falls to a minimum at \(i = e\) and then rises; both ends of the curve have the same deviation.
- Forgetting total reflection at the second face.Check \(r_2 = A - r_1\) against \(C\). If \(r_2 > C\), the ray does not emerge; if \(A > 2C\), no ray does.
- Using \(\delta = (\mu - 1)A\) for a \(60°\) prism.It holds only for a thin prism, a few degrees, near normal incidence.