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Most students can quote the minimum deviation formula. Far fewer can take a set of prism readings, draw a clean curve and explain why it has the shape it does. Examiners know this, which is why the i-δ graph turns up in viva questions and in experiment-based MCQs. This post walks through the practical with a full set of sample numbers, shows how to read the graph, and lists the errors that ruin a good curve.
Where is this practical in the syllabus?
The JEE (Main) 2026 syllabus lists, as experiment 14 in its experimental skills unit, "the plot of the angle of deviation vs angle of incidence for a triangular prism". The NEET (UG) 2026 syllabus carries the same line. The CBSE Physics curriculum for 2026-27 asks Class 12 students "to determine angle of minimum deviation for a given prism by plotting a graph between angle of incidence and angle of deviation". The theory is section 9.6 of the NCERT Class 12 chapter on ray optics (Reprint 2026-27). Check your own exam's bulletin each year.
Why does the i-δ graph dip and rise again?
NCERT derives two results for a prism of angle A: r₁ + r₂ = A, and δ = i + e − A. The second one is symmetric. Swap i and e and δ stays the same, because a light ray can retrace its path. So, as NCERT points out, any δ except the lowest one comes from two angles of incidence. The curve therefore falls, reaches a single minimum Dm where i = e, and rises again. At that point the ray inside travels parallel to the base, and
n = sin[(A + Dm)/2] / sin(A/2)
Worked example: a full set of sample readings
The numbers below are sample values, calculated from Snell's law for an ideal prism with A = 60° and n = 1.52 (NCERT's value for crown glass). Real readings will scatter by half a degree or so.
i = 35°, δ = 43.8°
i = 40°, δ = 40.6°
i = 45°, δ = 39.3°
i = 50°, δ = 38.9°
i = 55°, δ = 39.4°
i = 60°, δ = 40.5°
i = 65°, δ = 42.1°
Step 1: read the minimum. Draw a smooth curve, not straight segments. The lowest point sits at about Dm = 38.9°, near i = 49.5°.
Step 2: find n. (A + Dm)/2 = (60° + 38.9°)/2 = 49.45°. So n = sin 49.45° / sin 30° = 0.760/0.500 = 1.52, matching the glass we started with.
Step 3: check the symmetry. For i = 40°, the ray leaves at e = 60.6°. Since i + e = A + δ, the other angle that gives the same δ = 40.6° is i = 60.6°. Your curve should pass through both points at the same height.
Step 4: see how much precision matters. If you misread Dm as 38.5°, n comes out as 1.515. At 39.0° it is 1.521. Half a degree in Dm moves n by about 0.005, so spend your extra readings near the bottom of the curve.
Why do readings stop working at small angles of incidence?
At the second face the ray goes from glass to air, so it can be totally internally reflected. NCERT's table gives a critical angle of 41.14° for crown glass. For the ray to get out, r₂ = A − r₁ must stay below that, so r₁ must be more than about 18.9°. With n = 1.52 that needs sin i > 1.52 × sin 18.9°, which means i above about 29.4°. Start your table near 35°, not 20°.
Common mistakes and how to avoid them
Measuring the wrong angle. δ is the angle between the incident ray extended forward and the emergent ray traced back, not the angle between the emergent ray and the prism face.
Too few points near the bottom. Steps of 5° are fine on the arms. Use smaller steps around the minimum.
Drawing a V. Join points with a smooth curve. A sharp corner suggests a minimum that the physics does not allow.
Forgetting the prism angle. Measure A yourself rather than assuming 60°. Every n you calculate uses it.
A deviation below Dm. Any δ you calculate must be at least Dm for that prism. If you get less, recheck the angle.
For the formula side of the chapter, including thin prisms and dispersion, see our guide to prism problems and minimum deviation. If the reason the curve has two branches still feels abstract, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
A practice routine for this week
Day 1: derive δ = i + e − A from the NCERT figure without looking.
Day 2: take the sample table above, plot it on graph paper and read Dm and n yourself.
Day 3: recompute δ for i = 40° and i = 60° using Snell's law at both faces, then check the symmetry step.
Day 4: work out the smallest i that gets light out of a 60° prism with n = 1.5.
Day 5: write a five-line viva answer to "why is there only one minimum?"
Tonight, sketch the i-δ curve from memory and mark Dm, the two branches and the point where i = e.
