Geometry of the prism
Consider a triangular prism with refracting faces $AB$ and $AC$ meeting at the apex with refracting angle $A$. A ray $PQ$ strikes face $AB$ at angle of incidence $i$ and refracts into the glass at angle $r_1$. It travels to the second face $AC$, where it meets the surface at angle $r_2$ and emerges at the angle of emergence $e$. Because the medium is denser than air, the ray bends towards the normal on entry and away from the normal on exit, and the net effect is a turning of the ray towards the base of the prism.
The two normals at $Q$ and $R$, drawn inward, meet at a point $N$. In the quadrilateral $AQNR$ the angles at $Q$ and $R$ are right angles, so the remaining two angles sum to $180^\circ$. Within the internal triangle $QNR$, $r_1 + r_2 + \angle QNR = 180^\circ$. Comparing the two statements removes $\angle QNR$ and yields the prism's first key relation.
This is a purely geometric result: it does not depend on the refractive index, the wavelength of light or the angle of incidence. As long as the ray actually passes through both refracting faces, the two angles of refraction inside the glass must always add up to the refracting angle $A$. The relation is the hinge on which the entire prism analysis turns, because it lets every later equation be written in terms of measurable external angles ($i$, $e$, $A$) rather than the internal angles, which cannot be observed directly. NIOS §21.1.2 reaches the identical relation $r_1 + r_2 = A$ by using the fact that the interior angles of the quadrilateral $AQOR$ sum to $360^\circ$; the two derivations are equivalent statements of the same geometry.
Light bends towards the base; the deviation $\delta$ is the angle between the incident direction and the emergent ray (NCERT Fig. 9.21).
Prism angle relation: $\;A = r_1 + r_2$ — the refracting angle of the prism equals the sum of the two internal refraction angles.
The deviation relation
The total deviation $\delta$ is the angle between the original incident direction and the final emergent ray. It is built up from the bending at each face: a turn of $(i - r_1)$ at the first surface and a turn of $(e - r_2)$ at the second. Adding these and substituting $r_1 + r_2 = A$ collapses the internal angles out of the expression.
| Step | Statement | Origin |
|---|---|---|
| Deviation at each face | $\delta = (i - r_1) + (e - r_2)$ | Sum of the two turnings |
| Use $r_1 + r_2 = A$ | $\delta = i + e - A$ | Substitution |
| Equivalent form | $i + e = A + \delta$ | Rearrangement |
The relation $\delta = i + e - A$ is symmetric in $i$ and $e$: swapping the roles of the incident and emergent rays leaves $\delta$ unchanged. This mirrors the principle of reversibility of light — the ray could equally travel the path backwards, producing the same deviation. The symmetry has a direct consequence for the deviation curve discussed next.
It is worth pausing on what each term contributes. The angle of incidence $i$ and the angle of emergence $e$ are the two external angles the ray makes with the normals at the two faces; the refracting angle $A$ is fixed by the shape of the prism. Increasing $A$ for fixed $i$ and $e$ would reduce the deviation, but in practice $A$ is a property of the prism, so the deviation is controlled by the single adjustable quantity — the angle of incidence. Once $i$ is chosen, Snell's law at the first face fixes $r_1$, the relation $A = r_1 + r_2$ fixes $r_2$, and Snell's law at the second face fixes $e$; the deviation then follows. Every prism problem in NEET is, at heart, this chain of four steps applied in sequence.
Deviation versus incidence curve
If the angle of incidence is varied, the deviation first decreases, reaches a
single least value, then increases again. The plot of $\delta$ against $i$ is
therefore a U-shaped curve with one minimum. Because of the $i\leftrightarrow e$
symmetry, any value of $\delta$ above the minimum is produced by two distinct
angles of incidence — one where $i
The physical reason the curve has a minimum rather than falling without limit is the trade-off between the two faces. When the angle of incidence is small, the ray enters with little bending but strikes the second face at a large angle, producing a large emergence deviation; when the angle of incidence is large, the entry deviation is large but the exit bending is small. Somewhere between these extremes the two contributions balance, and the total deviation is least. This balance point is precisely the symmetric configuration, and it is the only configuration of practical importance for measuring refractive index, because it is the easiest to locate experimentally — one simply rotates the prism until the deviated image stops moving and begins to return.
A single deviation value (grey) is reached at two incidence angles; the minimum (red) occurs once, at $i = e$ (NCERT Fig. 9.22).
Minimum deviation and the prism formula
At the minimum, $\delta = D_m$ and $i = e$, which forces $r_1 = r_2$. The refracted ray inside the prism then runs parallel to the base, and the ray passes symmetrically through the glass. Setting $r_1 = r_2 = r$ in $A = r_1 + r_2$ gives $r = A/2$, while $\delta = i + e - A$ with $i = e$ gives $D_m = 2i - A$, hence $i = (A + D_m)/2$.
Applying Snell's law at the first face with these symmetric angles gives the prism formula, from which the refractive index follows once $A$ and $D_m$ are measured on a spectrometer.
At the symmetric position $i = (A + D_m)/2$ and $r = A/2$, so Snell's law $\sin i = n \sin r$ becomes $\sin[(A + D_m)/2] = n\,\sin(A/2)$, which rearranges directly to the boxed result. The strength of this formula is experimental: both $A$ and $D_m$ are angles that can be read off a spectrometer table to a fraction of a degree, and neither requires knowing the angle of incidence precisely — only that the prism has been set at minimum deviation. This makes the minimum-deviation method the standard laboratory technique for finding the refractive index of a transparent solid, and it is also why the prism formula is quoted only at minimum deviation rather than for a general angle of incidence.
$$n = \frac{\sin\!\left[\dfrac{A + D_m}{2}\right]}{\sin\!\left[\dfrac{A}{2}\right]}$$
The symmetric-ray conditions hold only at minimum deviation
The relations $i = e$, $r_1 = r_2 = A/2$ and "ray parallel to the base" are true only at the minimum-deviation position. For any other angle of incidence the ray is asymmetric ($r_1 \neq r_2$) and these shortcuts fail. Equally, do not use the thin-prism result $\delta = (n-1)A$ for a prism of large refracting angle — it is an approximation valid only for small $A$.
At minimum deviation only: $r_1 = r_2 = A/2$, $i = e$, internal ray $\parallel$ base. Thin prism only: $\delta = (n-1)A$.
A prism splits white light because $n$ depends on wavelength — see Dispersion by a Prism.
Thin-prism approximation
When the refracting angle $A$ is small, the minimum deviation $D_m$ is small too. In the prism formula the sines can then be replaced by their angles, so $n \approx (A + D_m)/A$, which rearranges to a compact result. The deviation of a thin prism is small and, to this order, independent of the angle of incidence — thin prisms do not bend light much, which is why they are used in pairs to cancel deviation while retaining dispersion.
Thin prism: $\;\delta = (n - 1)\,A$ — deviation is proportional to the refracting angle and to $(n-1)$.
NIOS §21.1.1–§21.1.2 develops the same geometry while emphasising dispersion: since $n$ is larger for violet than for red light, the deviation $\delta = (n-1)A$ is greatest for violet and least for red, so a beam of white light fans out into a spectrum on emerging from the prism.
The independence of $\delta$ from the angle of incidence in the thin-prism limit is the property that makes thin prisms practically useful. Because each thin prism deviates light by a fixed small amount set only by $(n-1)A$, two such prisms can be combined to engineer a desired optical behaviour. If a crown-glass prism and a flint-glass prism are arranged in opposition with their angles chosen so that the deviations cancel, the net deviation is zero while the difference in their dispersions survives — this is dispersion without deviation, the basis of the NEET 2017 question below. The reverse arrangement, deviation without dispersion, underlies the design of achromatic prisms used in optical instruments.
Prism in a denser medium
Every form of the prism formula uses the refractive index of the prism material relative to its surroundings. In air this is essentially the absolute index of the glass. If the same prism is immersed in water, the relevant index is the ratio of glass to water, which is smaller than the glass-to-air value. A smaller relative index produces a smaller deviation, so the angle of minimum deviation falls when the prism is moved from air into water.
The same idea explains a limiting case worth remembering: if a prism were placed in a medium of exactly its own refractive index, the relative index would be $1$, every refraction angle would equal its incidence angle, and the prism would produce no deviation at all — light would pass straight through as though the prism were not there. Water does not reach that limit for ordinary glass, but it moves the relative index part of the way towards it, which is why the deviation drops. For NEET, the safe working rule is to treat the $n$ in every prism equation as the index of the prism material relative to its surroundings, and to recompute it whenever the surrounding medium changes.
| Quantity | Relation | Reads as |
|---|---|---|
| Prism angle | $A = r_1 + r_2$ | Refracting angle from internal angles |
| Deviation | $\delta = i + e - A$ | Symmetric in $i$ and $e$ |
| Minimum deviation | $i = e,\; r_1 = r_2 = A/2$ | Symmetric passage, ray $\parallel$ base |
| Refractive index | $n = \dfrac{\sin[(A+D_m)/2]}{\sin(A/2)}$ | Prism formula (spectrometer method) |
| Thin prism | $\delta = (n-1)A$ | Small $A$, independent of $i$ |
Worked examples
A prism of refracting angle $A = 60^\circ$ has an angle of minimum deviation $D_m = 40^\circ$ for a parallel beam of light. Find the refractive index of the prism material. Predict (qualitatively) the new minimum deviation if the prism is placed in water of index $1.33$.
Apply the prism formula: $n = \dfrac{\sin[(A + D_m)/2]}{\sin(A/2)} = \dfrac{\sin 50^\circ}{\sin 30^\circ} = \dfrac{0.766}{0.5} \approx 1.532.$
In water the relative index becomes $n_{\text{rel}} = 1.532/1.33 \approx 1.152$. Since the prism formula now uses this smaller index, the angle of minimum deviation in water is markedly less than $40^\circ$ (a numerical solve of the prism formula with $n_{\text{rel}}$ and $A = 60^\circ$ gives a much smaller $D_m$).
A thin prism of refracting angle $6^\circ$ is made of glass of refractive index $1.5$. What deviation does it produce?
For a thin prism $\delta = (n-1)A = (1.5 - 1)(6^\circ) = 0.5 \times 6^\circ = 3^\circ$. The deviation is independent of the (small) angle of incidence.
Refraction Through a Prism in one screen
- Prism angle equals the sum of internal refraction angles: $A = r_1 + r_2$.
- Deviation: $\delta = i + e - A$, symmetric in $i$ and $e$ (reversibility).
- The $\delta$–$i$ curve is U-shaped; one $\delta$ → two values of $i$, except at the minimum.
- At minimum deviation: $i = e$, $r_1 = r_2 = A/2$, internal ray parallel to base.
- Prism formula: $n = \dfrac{\sin[(A+D_m)/2]}{\sin(A/2)}$ — the spectrometer method for $n$.
- Thin prism: $\delta = (n-1)A$, small and independent of incidence angle.
- In a denser surrounding medium the relative index drops, so $D_m$ decreases.