ARCHETYPE: Prism, Deviation & Dispersion Numericals

The minimum-deviation formula is not a museum piece — it is how the refractive index of an optical glass is measured, to five decimal places, on a spectrometer, by reading the least deviation of a prism cut from it. Every catalogue value of $\mu$ began as an $A$ and a $\delta_m$. The dispersive-power balance is the working principle of the achromatic lens in every camera, microscope and pair of binoculars: a crown element and a flint element cemented together so their colour spreads cancel while their focusing powers do not, exactly equation (6) applied to lenses instead of prisms — the invention that made sharp, colour-free optics possible, which Chester Moore Hall first devised around 1730 and John Dollond patented in 1758, its priority contested in the courts in the 1760s. The direct-vision prism lives in pocket spectroscopes and hand-held mineral identifiers, letting you see an emission spectrum straight down the barrel. And the same dispersion, spread across a diffraction grating's successor, is how a spectrometer splits starlight or a chemical flame into the lines that reveal what it is made of — the science Fraunhofer's solar lines began.

A prism is a wedge of glass with a refracting angle (apex angle) $A$. A ray entering one face is refracted toward the base, crosses to the second face, and is refracted again on the way out, emerging turned through a total angle of deviation $\delta$ from its original path. Track the angles: the ray refracts from incidence $i$ to an internal angle $r_1$ at the first face, travels to the second face where it makes an internal angle $r_2$ and emerges at angle $e$. Two facts of geometry govern everything. The internal angles fill the apex, $r_1 + r_2 = A$, and the total deviation is the sum of the two individual turns, $\delta = (i - r_1) + (e - r_2) = i + e - A$. As you rotate the prism, $\delta$ first falls, reaches a minimum, then rises again — and the minimum occurs exactly when the passage is symmetric, $i = e$ and $r_1 = r_2 = A/2$. That symmetry is what makes the minimum-deviation measurement so clean: it needs no knowledge of the internal path, only the apex angle and the least deviation you can read off a spectrometer. At that special geometry, $\delta_m = 2i - A$, and Snell's law at the first face (quoted from earlier in this chapter) gives the master relation the whole topic hangs on.

Here is the master recipe for every problem below. First, name the quantity asked — an index, a deviation, a dispersive power, or a combination angle. Second, pick the matching relation from the four. Third, for a combination, set the crown-against-flint balance — deviations equal for direct-vision, dispersions equal for achromatic. Fourth, check the apex limit $A \le 2C$, because a prism whose apex is too wide throws the internal ray past the critical angle at the second face, where it total-internally-reflects instead of emerging, and no answer exists.

Everything numerical about a prism is in these four lines. At minimum deviation,

where: $A$ is the refracting angle, $\delta_m$ the minimum deviation and $\mu$ the refractive index. For a thin prism (small $A$, so all angles small and sines equal their arguments), this collapses to the linear

independent of the angle of incidence — the deviation is just the "extra bending power" $(\mu - 1)$ times the apex. Because $\mu$ is larger for violet than for red, each colour gets its own deviation, and the spread between the extremes is the angular dispersion

while the colour-spread measured per unit mean deviation is the dispersive power

where: $\mu_v$, $\mu_r$, $\mu_y$ are the indices for violet, red and mean (yellow) light. Note what (4) does not contain: the apex $A$ cancels, so dispersive power is a pure property of the glass, dimensionless, the same for a fat prism or a thin one. Crown glass runs around $\omega = 0.017$, flint glass roughly double — the difference that makes achromatic optics possible. These four relations, plus the two combination balances, answer every prism numerical there is.

Put two thin prisms base-to-base of opposite orientation — a crown prism of angle $A$ and a flint prism of angle $A'$ — and their deviations and dispersions each subtract. Two designs matter. A direct-vision prism spreads colour but leaves the mean beam undeviated, which needs the mean deviations to cancel:

An achromatic combination bends the beam but leaves no residual colour, which needs the dispersions to cancel:

Equation (5) is the spectroscope's direct-vision prism (see the spectrum straight ahead); equation (6) is the achromatic lens's secret (focus all colours together). Every combination problem is one of these two balances solved for the unknown flint angle $A'$.

1 · Board level — refractive index from minimum deviation (CBSE and NEET pattern)

A prism of refracting angle $60^\circ$ produces a minimum deviation of $40^\circ$. Find the refractive index of its glass.

Step 1. Recognise the type: given $A$ and $\delta_m$, asked for $\mu$ — this is equation (1) used forward.

Step 2. Substitute: $\mu = \dfrac{\sin\!\big(\tfrac{60 + 40}{2}\big)}{\sin\!\big(\tfrac{60}{2}\big)} = \dfrac{\sin 50^\circ}{\sin 30^\circ}$.

Step 3. Evaluate: $\sin 50^\circ = 0.766$, $\sin 30^\circ = 0.5$, so $\mu = \dfrac{0.766}{0.5} = \boxed{1.532}$. A typical dense-crown or light-flint glass.

Step 4. Sanity check: $\mu$ should exceed $1$ (glass is denser than air) and lie in the $1.4$–$1.7$ range for common glasses — $1.53$ sits comfortably there. Had you written $\sin(40/2)$ on top you would have got $\sin 20^\circ/\sin 30^\circ = 0.684$, a "refractive index" below one — the instant tell of the half-sum error.

2 · Board level — minimum deviation and a thin prism of crown glass (CBSE and ICSE pattern)

(a) Find the minimum deviation of a $60^\circ$ crown-glass prism of refractive index $1.5$, and the angle of incidence at which it occurs. (b) Find the deviation produced by a thin $5^\circ$ crown prism of the same glass.

Step 1. Part (a), invert equation (1) for $\delta_m$: $\sin\!\big(\tfrac{60 + \delta_m}{2}\big) = \mu\,\sin\!\big(\tfrac{60}{2}\big) = 1.5 \times \sin 30^\circ = 1.5 \times 0.5 = 0.75$.

Step 2. Take the inverse sine: $\dfrac{60 + \delta_m}{2} = \sin^{-1}(0.75) = 48.59^\circ$, so $60 + \delta_m = 97.18^\circ$ and $\delta_m = \boxed{37.2^\circ}$. The angle of incidence at minimum deviation is $i = \dfrac{A + \delta_m}{2} = 48.59^\circ$, and the internal angle is $r = A/2 = 30^\circ$.

Step 3. Part (b), a thin prism, so use equation (2): $\delta = (\mu - 1)A = (1.5 - 1)\times 5^\circ = 0.5 \times 5 = \boxed{2.5^\circ}$.

Step 4. Note the two regimes side by side: the $60^\circ$ prism needs the full trigonometric formula (its angles are large), while the $5^\circ$ prism uses the linear shortcut — and you must never apply $(\mu - 1)A$ to the $60^\circ$ prism, which would give a wrong $30^\circ$ instead of $37.2^\circ$.

3 · Competitive level — a direct-vision combination (JEE-Main and NEET pattern)

A crown-glass prism of refracting angle $6^\circ$ and mean index $1.517$ is combined with a flint prism of mean index $1.65$ so that the combination produces dispersion but no net mean deviation (a direct-vision prism). Find the flint prism's refracting angle.

Step 1. Direct-vision means zero net mean deviation, equation (5): $(\mu_y - 1)A = (\mu'_y - 1)A'$, the crown's deviation balanced by the flint's.

Step 2. Solve for the flint angle: $A' = \dfrac{(\mu_y - 1)A}{\mu'_y - 1} = \dfrac{(1.517 - 1)\times 6}{1.65 - 1} = \dfrac{0.517 \times 6}{0.65}$.

Step 3. Evaluate: $A' = \dfrac{3.102}{0.65} = \boxed{4.77^\circ}$. The flint prism is smaller than the crown because flint bends more strongly per degree of apex.

Step 4. Confirm the colour survives: the two dispersions do not cancel (they would need equation (6), a different balance), so white light still fans into a spectrum — seen straight ahead, which is the whole purpose of a direct-vision prism in a pocket spectroscope. Timing note: under two minutes. The only decision is which balance to use — deviations for direct-vision, dispersions for achromatic — and once chosen it is a single division. Write the balance equation before you touch the calculator.

4 · Competitive level — dispersive power and an achromatic pair (JEE-Advanced and NEET pattern)

Crown glass has $\mu_r = 1.514$, $\mu_y = 1.517$, $\mu_v = 1.523$. (a) Find its dispersive power. (b) A $6^\circ$ crown prism is to be made achromatic (no net dispersion) with a flint prism whose $(\mu'_v - \mu'_r) = 0.020$. Find the flint prism's refracting angle.

Step 1. Part (a), dispersive power, equation (4): $\omega = \dfrac{\mu_v - \mu_r}{\mu_y - 1} = \dfrac{1.523 - 1.514}{1.517 - 1} = \dfrac{0.009}{0.517} = \boxed{0.0174}$ — small and dimensionless, a property of crown glass alone.

Step 2. Part (b), achromatic means zero net dispersion, equation (6): $(\mu_v - \mu_r)A = (\mu'_v - \mu'_r)A'$, the crown's dispersion balanced by the flint's.

Step 3. Solve: $A' = \dfrac{(\mu_v - \mu_r)A}{\mu'_v - \mu'_r} = \dfrac{0.009 \times 6}{0.020} = \dfrac{0.054}{0.020} = \boxed{2.7^\circ}$.

Step 4. Read the physics: because the flint disperses more strongly, a smaller flint prism ($2.7^\circ$) cancels the crown's colour spread — but leaves a net deviation (the mean beam is bent), which is exactly what makes an achromatic combination useful in a lens: colours focus together, and the beam is still steered. The crown's own angular dispersion here is $(\mu_v - \mu_r)A = 0.009 \times 6 = 0.054^\circ$. Timing note: about two minutes for both parts. The dispersive power is one subtraction over another; the achromat is one balance solved for $A'$. Keep crown and flint quantities in separate columns so you never balance a crown deviation against a flint dispersion.

The half-sum over the half-apex. Equation (1) is $\sin\!\big(\tfrac{A+\delta_m}{2}\big)$ over $\sin\!\big(\tfrac{A}{2}\big)$. Writing $\sin(\delta_m/2)$, or failing to halve, is the number-one lost mark — it produces a "refractive index" below one. Anchor it: at minimum deviation the incidence angle is $(A+\delta_m)/2$ and the internal angle is $A/2$.

Thin-prism formula on a fat prism. $\delta = (\mu - 1)A$ holds only for small $A$. Apply it to a $60^\circ$ prism and you get $30^\circ$ instead of the correct $37.2^\circ$. Use the full formula whenever $A$ is more than a few degrees.

Balancing the wrong quantity in a combination. Direct-vision (no net bend) balances the deviations $(\mu_y - 1)A$; achromatic (no net colour) balances the dispersions $(\mu_v - \mu_r)A$. Swapping the two gives a plausible-looking but wrong flint angle.

Getting the colour order backwards. Violet has the higher index, so violet deviates more than red. A spectrum drawn with red bent most is wrong, and any "$\mu_v < \mu_r$" is a red flag.

Ignoring the apex limit. A prism with $A > 2C$ transmits nothing — the internal ray total-internally-reflects at the second face. For $\mu = 1.5$, $C = 41.8^\circ$, so $A$ must stay below $83.6^\circ$. Always sanity-check before quoting a deviation.

Board scope. Prism deviation-and-dispersion numericals are examined by CBSE, ICSE, JEE-Main, JEE-Advanced and NEET. IGCSE 0625 treats dispersion by a prism only qualitatively (white light into the seven colours) and sets no minimum-deviation or dispersive-power calculations. AP Physics 2 carries none of this, and IB and A-Level have no prism optics. If your board is one of those, this gym is background — learn the qualitative spectrum instead.

Texts. NCERT Physics Class 12, Ch. 9 ("Ray Optics and Optical Instruments"), refraction through a prism, dispersion and the dispersive-power combinations · H. C. Verma, Concepts of Physics Vol. 1, "Dispersion and Spectra" · D. C. Pandey, Optics and Modern Physics, the prism problem set for JEE and NEET.

Provenance. Standard optics. Recompute: $\mu = \sin 50^\circ/\sin 30^\circ = 1.532$; $\delta_m = 37.2^\circ$ (with $i = 48.59^\circ$, $r = 30^\circ$) from $A = 60^\circ$, $\mu = 1.5$; thin-prism $\delta = (1.5 - 1)\times 5 = 2.5^\circ$; dispersive power $\omega = 0.009/0.517 = 0.0174$; angular dispersion $(\mu_v - \mu_r)\times 6 = 0.054^\circ$; direct-vision $A' = 0.517 \times 6/0.65 = 4.77^\circ$; achromatic $A' = 0.009 \times 6/0.020 = 2.7^\circ$; the grazing-limit index $\mu = \sqrt{3} = 1.732$ when $\delta_m = A = 60^\circ$; and the apex limit $A \le 2C = 2\sin^{-1}(1/1.5) = 83.6^\circ$.

Prerequisites. The Prism: Deviation, Dispersion and Dispersive Power — the concept article that derives these relations, which this gym drills. Snell's law and the critical angle were established earlier in this chapter and are quoted here. Leads to. Instrument Magnification Numericals — the sibling gym for the optical instruments · the university Dispersion by Prisms seam.

Go deeper (UG). The wavelength-dependence of refractive index itself — the Cauchy and Sellmeier dispersion formulas that give $\mu(\lambda)$, and the physics of why glass disperses — is developed in Dispersion by Prisms and Chromatic Aberration and Achromats in the university optics track. Nothing here needs it; the prism relations are pure ray geometry plus a quoted Snell's law.