Trace incident and emergent rays through a triangular prism for several incidence angles i. Plot deviation δ vs i — the curve has a minimum δ_min. At minimum deviation, n = sin((A + δ_min)/2) / sin(A/2). Refractive index n is independent of i at minimum.
-- NCERT Physics Lab Manual Class 12, Part 9, p. 99Experiment 14 Prism Angle of Deviation
8 MCQs9-step worked example
Source: NCERT Experimental SkillsOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Experiment 14 Prism Angle of Deviation, explained for NEET
The usual failure here is treating the i–δ graph as a straight line with a lowest point at one end. It is not. As the angle of incidence is increased from small values, the deviation first falls, reaches a single minimum, and then rises again. A student who plots four points on the rising side only, joins them, and reads the smallest plotted δ as the minimum deviation reports a value that is too large — and the refractive index that follows is wrong in the same direction.
The experiment is the one set out in the NCERT Physics Lab Manual Class 12, Part 9, page 99: a triangular glass prism is mounted, rays are traced for a spread of incidence angles, and δ is measured for each. Two relations carry the whole analysis. For any ray through the prism, A + δ = i + e, where A is the refracting angle, i the incidence angle and e the emergence angle. At the minimum, and only there, i = e, the ray inside the prism runs parallel to the base, and δ takes the value δ_m. The refractive index then follows from μ = sin[(A + δ_m)/2] / sin(A/2).
Two consequences are worth holding on to. First, the curve is flat near its bottom, so several nearby incidence angles give almost the same deviation — that is why the graphical method is used rather than a single reading. Second, because the curve is U-shaped, any δ above δ_m is produced by two different angles of incidence, one on each side of the minimum; these are the i and e of the same ray, swapped.
Watch out at the reporting stage. Read δ_m off the smooth curve, not off the nearest plotted dot, and quote it to the precision your protractor actually allows — not to the digits the calculator prints.
Can you answer these Experiment 14 Prism Angle of Deviation MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
As the angle of incidence on a triangular prism is increased steadily from small values, the angle of deviation:
Show answer and why every option is right or wrong
Answer: C. Option C is correct: the i–δ plot for a prism is a U-shaped curve with one minimum, which is the whole reason the graph is plotted rather than a single reading taken (NCERT Physics Lab Manual Class 12, Part 9, page 99).
Why A is wrong: A is wrong because a monotonic rise would leave no minimum to locate, and the experiment exists precisely to locate one.
Why B is wrong: B is wrong because the fall stops at the minimum; beyond it the deviation rises again as i approaches grazing values.
Why D is wrong: D is wrong because deviation varies continuously with i throughout the usable range; the curve is never flat over an extended stretch.
MCQ 2Easy RecallPractice
At the position of minimum deviation for a triangular prism, the angle of incidence and the angle of emergence are related how?
Show answer and why every option is right or wrong
Answer: A. Option A is correct: minimum deviation occurs in the symmetric configuration, i = e, with the ray inside the prism parallel to the base (NCERT Physics Lab Manual Class 12, Part 9, page 99).
Why B is wrong: B is wrong because no factor of two appears anywhere in the symmetric condition; it is a plain equality.
Why C is wrong: C is wrong because it misquotes the prism relation — i + e equals A + δ, not A alone.
Why D is wrong: D is wrong because a zero emergence angle would mean the ray leaves along the normal, which is not the symmetric minimum-deviation path.
MCQ 3Easy RecallPractice
For a ray passing through a triangular prism of refracting angle A, with incidence angle i, emergence angle e and deviation δ, the correct relation is:
Show answer and why every option is right or wrong
Answer: D. Option D is correct: A + δ = i + e holds for every ray through the prism, not only at minimum deviation, and it is the relation used to check each row of the observation table (NCERT Physics Lab Manual Class 12, Part 9, page 99).
Why A is wrong: A is wrong because it puts the deviation on the same side as the two ray angles; deviation is an output of i and e, not a third input summed with them.
Why B is wrong: B is wrong because the relation involves the sum of i and e, never their difference.
Why C is wrong: C is wrong because it flips the sign of δ, which would make the deviation shrink as the ray angles grow.
MCQ 4Direct ApplicationPractice
A prism has a refracting angle of 60° (exact) and its measured minimum deviation is 30.0°. The angle of incidence at which this minimum occurs is:
Show answer and why every option is right or wrong
Answer: B. Option B is correct: at minimum deviation i = e, so A + δ_m = 2i gives i = (60° + 30.0°)/2 = 45.0° (NCERT Physics Lab Manual Class 12, Part 9, page 99).
Why A is wrong: A is wrong because it simply repeats the deviation value instead of halving the sum A + δ_m.
Why C is wrong: C is wrong because it repeats the refracting angle; the incidence angle equals A only if δ_m also equalled A.
Why D is wrong: D is wrong because it comes from adding A and δ_m without halving — grazing incidence is not the minimum-deviation setting.
MCQ 5Direct ApplicationPractice
In one row of the observation table for a prism of refracting angle 60° (exact), the incidence angle is 50.0° and the emergence angle is 40.0°. The angle of deviation for that ray is:
Show answer and why every option is right or wrong
Answer: D. Option D is correct: δ = i + e − A = 50.0° + 40.0° − 60° = 30.0° (NCERT Physics Lab Manual Class 12, Part 9, page 99).
Why A is wrong: A is wrong because it is the difference i − e; the relation uses the sum of the two ray angles.
Why B is wrong: B is wrong because it halves the correct result, as if the symmetric i = e condition had been imposed on a row where it does not hold.
Why C is wrong: C is wrong because it averages i and e, (50.0° + 40.0°)/2 = 45.0°, and forgets to subtract the prism angle; the deviation is i + e − A.
MCQ 6Direct ApplicationPractice
Deviation angles in this experiment are read on a protractor whose smallest division is 0.5°. A calculator average of several readings gives 38.2462°. The value that should be entered in the table is:
Show answer and why every option is right or wrong
Answer: A. Option A is correct: the reported value cannot be finer than the instrument's smallest division, so the average rounds to the nearest half-degree. 38.2462° lies 0.246° from 38.0° and 0.254° from 38.5°, so it rounds to 38.0° (NCERT Physics Lab Manual Class 12, Part 9, page 99).
Why B is wrong: B is wrong because 0.25° is half a scale division — a precision the protractor cannot deliver, however the arithmetic came out.
Why C is wrong: C is wrong because three decimal places claim a resolution roughly 500 times finer than the instrument has.
Why D is wrong: D is wrong because it copies the calculator output unchanged, which is the classic over-reporting error in this experiment's table.
MCQ 7CalculationPractice
A smooth i–δ curve for a prism of refracting angle 60° (exact) has its lowest point at a deviation of 38.0°. The refractive index of the prism material is closest to: (take sin 49.0° = 0.755)
Show answer and why every option is right or wrong
Answer: C. Option C is correct: (A + δ_m)/2 = 49.0° and A/2 = 30.0°, so μ = sin 49.0° / sin 30.0° = 0.755/0.500 = 1.51 (NCERT Physics Lab Manual Class 12, Part 9, page 99).
Why A is wrong: A is wrong because it is the familiar value for water, reached only by ignoring the measured δ_m entirely.
Why B is wrong: B is wrong because it is what the same formula gives for δ_m = 30.0°, i.e. by reading the minimum off the wrong point of the curve.
Why D is wrong: D is wrong because it comes from a sine ratio evaluated at the wrong half-angle: sin 60.0°/sin 30.0° = 1.73 uses A in place of (A + δ_m)/2 in the numerator.
MCQ 8Concept TrapPractice
A student inspects the completed i–δ curve and picks a deviation value clearly larger than the minimum. The number of distinct angles of incidence that produce that deviation is:
Show answer and why every option is right or wrong
Answer: B. Option B is correct: a horizontal line above the minimum cuts the U-shaped curve twice, and the two incidence angles are the i and e of the same ray with their roles interchanged (NCERT Physics Lab Manual Class 12, Part 9, page 99).
Why A is wrong: A is wrong because a single incidence angle corresponds only to the minimum deviation itself, where the two branches meet.
Why C is wrong: C is wrong because the curve has one minimum and two branches, so a horizontal line can meet it at most twice.
Why D is wrong: D is wrong because it would require the curve to oscillate; the measured plot falls once and rises once.
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How do you solve a Experiment 14 Prism Angle of Deviation question? A worked example
- 1
Given.
Refracting angle of the prism A = 60° (exact, ruled prism). The i–δ curve is plotted from readings taken with a protractor of smallest division 0.5°, and its lowest point lies at δ_m = 38.0°. Use sin 49.0° = 0.755 and sin 30.0° = 0.500 (exact).
- 2
Required.
The refractive index μ of the prism material, and the incidence angle at which the minimum occurs.
- 3
Concept.
At minimum deviation the path through the prism is symmetric: i = e and the internal ray is parallel to the base. The general relation A + δ = i + e then collapses to A + δ_m = 2i, and the refractive index follows from the minimum-deviation relation given in the lab manual procedure (NCERT Physics Lab Manual Class 12, Part 9, page 99).
- 4
Formula.
A + δ_m = 2i, and μ = sin[(A + δ_m)/2] / sin(A/2).
- 5
Substitution.
(A + δ_m)/2 = (60° + 38.0°)/2 = 49.0°; A/2 = 60°/2 = 30.0°. So μ = sin 49.0° / sin 30.0° = 0.755 / 0.500.
- 6
Calculation.
μ = 1.510. The incidence angle at the minimum is i = 49.0°. The refracting angle 60° is an exact ruled value and the factor 2 in (A + δ_m)/2 is a counting integer; neither contributes to the significant-figure count, so the precision is fixed by δ_m, read to the protractor's 0.5° resolution — three significant figures.
- 7
Final answer.
μ = 1.51, with the minimum occurring at an incidence angle of 49.0°.
- 8
Common trap.
Reading δ_m as the smallest plotted value rather than the lowest point of the smooth curve. Plotted points rarely land exactly on the minimum, so the smallest dot sits above it; the resulting δ_m is too large and μ comes out too high. Draw the free-hand smooth curve first, then drop a perpendicular from its lowest point to the δ axis.
- 9
Similar NEET-style question.
For the same prism, a horizontal line drawn across the curve at δ = 45.0° meets it at i = 35.0° and at a second incidence angle. Using A + δ = i + e, find that second angle. (Answer: e = 45.0° + 60° − 35.0° = 70.0°, so the second incidence angle is 70.0°.)
What to remember before solving Experiment 14 Prism Angle of Deviation questions
More in Experimental Skills: 6 exam traps and mistakes · 3 formulas · 1 question pattern from its other lessons.
Experiment 14 Prism Angle of Deviation questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
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