Prism

Q: What is the angle of deviation of a prism?

The angle of deviation is the angle between the direction of the incident ray and the emergent ray. For a prism of angle A, it equals i plus e minus A, where i is the angle of incidence and e the angle of emergence. The ray always bends towards the base of the prism.

Q: What is the condition for minimum deviation in a prism?

Deviation is minimum when the ray passes symmetrically: the angle of incidence equals the angle of emergence and the angles inside the prism are both A over 2. In an isosceles prism the ray then travels parallel to the base. There is only one angle of incidence that gives minimum deviation.

Q: How is the refractive index of a prism found from minimum deviation?

At minimum deviation r equals A over 2 and i equals A plus delta m over 2. Snell's law then gives n equals sin of half of A plus delta m divided by sin of half of A. Measuring delta m on a spectrometer gives the refractive index of the glass accurately.

Q: Why does a thin prism deviate light by (n - 1)A?

For a thin prism and small angle of incidence all angles are small, so sines can be replaced by the angles themselves. Then i equals n r1 and e equals n r2, and the deviation i plus e minus A becomes n times A minus A, that is (n - 1)A, independent of the angle of incidence.

Q: Why does a prism split white light into colours?

The refractive index of glass depends slightly on wavelength: it is larger for violet than for red. Each colour is therefore deviated by a different amount, violet the most and red the least, and white light spreads into a spectrum. This is called dispersion.

Q: What is dispersive power?

Dispersive power is the ratio of angular dispersion to mean deviation, equal to n violet minus n red divided by n yellow minus 1. It depends only on the material, not on the prism angle. Flint glass has roughly twice the dispersive power of crown glass.

Q: Is the prism important for JEE Main and JEE Advanced?

Yes. JEE Main regularly asks minimum deviation, the refractive index formula, thin prism deviation and the deviation against incidence graph. JEE Advanced adds grazing incidence and emergence, conditions for no emergence using the critical angle, and achromatic or direct vision prism combinations.

Q: Which prism questions come in NEET?

NEET mostly asks the prism formula at minimum deviation, thin prism deviation (n - 1)A, the relation r1 plus r2 equals A, and conceptual questions on the deviation graph and on why violet light deviates most. Practise finding n from the angle of minimum deviation.

PhysicsRay Optics And Optical InstrumentsFor JEE aspirants

A prism is a transparent block with two plane refracting faces inclined at an angle $A$ . Light passing through a prism is bent towards its base by the deviation $δ = i + e - A$ , which is smallest when the path is symmetric; then $n = \frac{sin [( A + δ _{m} ) /2 ]}{sin ( A /2 )}$ . Because $n$ depends on colour, a prism also splits white light into a spectrum. Refraction and dispersion by a prism are core topics for JEE Main, JEE Advanced and NEET.

On this page1Refraction through a prism2Deviation graph3Minimum deviation4Limiting cases5Thin prism6Dispersion7Solved examples

Key Formulas - Quick Reference

★ Must learnGeometry: $r_{1} + r_{2} = A$ ; deviation $δ = i + e - A$
★ Must learnMinimum deviation ( $i = e$ , $r_{1} = r_{2} = \frac{A}{2}$ ): $n = \frac{sin [( A + δ _{m} ) /2 ]}{sin ( A /2 )}$
★ Must learnThin prism: $δ = (n - 1) A$
Emergence: all rays leave if $A < C$ ; none leave if $A > 2 C$
Grazing emergence: $sin i_{m i n} = n sin (A - C)$ ; $δ_{m a x} = i_{m i n} + 9 0^{\circ} - A$
★ Must learnAngular dispersion $θ = δ_{v} - δ_{r} = (n_{v} - n_{r}) A$ ; dispersive power $ω = \frac{n _{v} - n _{r}}{n _{y} - 1}$
Achromatic pair: $ω δ + ω^{'} δ^{'} = 0$ ; direct vision: $(n_{y} - 1) A = (n_{y}^{'} - 1) A^{'}$

1. Refraction Through a Prism

A prism is a homogeneous transparent medium bounded by two plane faces inclined at an angle $A$ , the angle of the prism (refracting angle). The third face is the base. A ray entering one face is refracted twice and emerges bent towards the base.

Figure 1: Path through a prism (

n = 1.5

A = 6 0^{\circ}

i = 4 5^{\circ}

). Refraction at both faces bends the ray towards the base:

r_{1} = 28. 1^{\circ}

r_{2} = 31. 9^{\circ}

e = 52. 4^{\circ}

δ = i + e - A = 37. 4^{\circ}

First face: $sin i = n sin r_{1}$ . Second face: $n sin r_{2} = sin e$ .
The normals at the two faces meet at $N$ , making $∠ QN R = 18 0^{\circ} - A$ (Figure 2). In triangle $QN R$ : $r_{1} + r_{2} = A$ .
Deviation at the first face $= i - r_{1}$ , at the second $= e - r_{2}$ . Adding: $δ = (i + e) - (r_{1} + r_{2}) = i + e - A$ .

Figure 2: The normals at

Q

and

R

meet at

N

with

∠ QN R = 18 0^{\circ} - A

(the quadrilateral

A QN R

has two right angles). Triangle

QN R

then gives

r_{1} + r_{2} = A

(here

28. 1^{\circ} + 31. 9^{\circ} = 6 0^{\circ}

Prism relations:

r_{1} + r_{2} = A

and

δ = i + e - A

. Because light paths are reversible, swapping

i

and

e

gives the same deviation. Use the prism relations when a ray crosses two faces inclined to each other; use slab relations for parallel faces.

Key idea

Every prism problem runs on three equations:

sin i = n sin r_{1}

r_{1} + r_{2} = A

n sin r_{2} = sin e

2. Deviation Against Angle of Incidence

As $i$ increases from the smallest value that lets light out, the deviation first falls rapidly, reaches a minimum $δ_{m}$ and then rises slowly.

Figure 3:

δ

against

i

(

A = 6 0^{\circ}

n = 1.5

), traced exactly. Minimum

δ_{m} = 37. 2^{\circ}

i = e = 48. 6^{\circ}

. Every other deviation occurs twice, at

i

and at the matching

e

(for

4 5^{\circ}

31. 2^{\circ}

and

73. 8^{\circ}

). The rise to the right is slower than the fall on the left.

For every $δ > δ_{m}$ there are two angles of incidence; they are each other's angles of emergence.
There is exactly one angle of incidence that gives the minimum deviation.
The curve is not symmetric: the right branch (large $i$ ) rises more slowly than the left branch falls.

3. Minimum Deviation

Figure 4: Minimum deviation. The path is symmetric:

i = e

r_{1} = r_{2} = \frac{A}{2}

, and inside an isosceles prism the ray runs parallel to the base. Here

δ_{m} = 37. 2^{\circ}

, so

n = \frac{sin [( A + δ _{m} ) /2 ]}{sin ( A /2 )} = 1.5

At minimum deviation the path is symmetric: $i = e$ , and then $r_{1} = r_{2} = r$ .
From $r_{1} + r_{2} = A$ : $r = \frac{A}{2}$ . From $δ_{m} = 2 i - A$ : $i = \frac{A + δ _{m}}{2}$ .
Snell's law at the first face:
$n = \frac{sin i}{sin r} = \frac{sin \frac{A + δ _{m}}{2}}{sin \frac{A}{2}}$

Measuring $δ_{m}$ with a spectrometer is the standard laboratory way to find the refractive index of glass. If the prism sits in a medium of index $n_{s}$ , the left side becomes $\frac{n}{n _{s}}$ .

Exam Trick

Equilateral prism with $δ_{m} = A = 6 0^{\circ}$ means $n = 3$ . Then $i = 6 0^{\circ}$ and $n = \frac{sin 6 0 ^{\circ}}{sin 3 0 ^{\circ}}$ . More generally, "ray passes symmetrically" or "ray parallel to the base" in an isosceles prism both mean minimum deviation.

Quick Recall: tap to check

At minimum deviation, what are

r_{1}

and

r_{2}

for a prism of angle

A

Both equal

\frac{A}{2}

6 0^{\circ}

prism has

δ_{m} = 3 0^{\circ}

. Find

n

n = \frac{sin 4 5 ^{\circ}}{sin 3 0 ^{\circ}} = 2

Why can two different angles of incidence give the same deviation?

Reversibility: the path with incidence

e

and emergence

i

has the same

i + e

, so the same

δ

Key idea

Minimum deviation means a symmetric path:

r = \frac{A}{2}

i = \frac{A + δ _{m}}{2}

, and

n

follows from Snell's law.

4. Limiting Cases: When Can Light Leave the Prism?

At the second face light goes from glass to air, so it can be totally reflected. With $r_{1} \leq C$ always (even for grazing incidence) and $r_{2} = A - r_{1}$ :

$A > 2 C$ : $r_{2} \geq A - C > C$ for every ray. No light emerges from the second face.
$A \leq C$ : $r_{2} \leq A \leq C$ . Every ray that enters emerges.
$C < A \leq 2 C$ : rays emerge only if $i$ exceeds $i_{m i n}$ , given by grazing emergence ( $e = 9 0^{\circ}$ , $r_{2} = C$ ): $sin i_{m i n} = n sin (A - C)$ .

Figure 5: Emergence depends on

A

versus

C

(exact traces,

n = 1.5

C = 41. 8^{\circ}

A < C

: every ray leaves.

C < A < 2 C

: only rays with

i

above

i_{m i n}

leave.

A > 2 C = 83. 6^{\circ}

: no ray can leave the second face.

Figure 6: The two ends of the

δ

i

curve (

A = 6 0^{\circ}

n = 1.5

) are reversed paths of each other: grazing incidence (

r_{1} = C

) and grazing emergence (

r_{2} = C

) both give

δ_{m a x} = 9 0^{\circ} + 27. 9^{\circ} - 6 0^{\circ} = 57. 9^{\circ}

. Below

i = 27. 9^{\circ}

the ray is totally reflected at the second face.

Maximum deviation occurs at the two ends of the curve, grazing incidence ( $i = 9 0^{\circ}$ ) or grazing emergence ( $e = 9 0^{\circ}$ ), which are reversed paths of each other:

δ_{m a x} = i_{m i n} + 9 0^{\circ} - A

JEE Advanced

Maximum and minimum deviation in one go. For $A = 6 0^{\circ}$ , $n = 1.5$ : $C = 41. 8^{\circ}$ , $A - C = 18. 2^{\circ}$ , $sin i_{m i n} = 1.5 sin 18. 2^{\circ} = 0.468$ , so $i_{m i n} = 27. 9^{\circ}$ and $δ_{m a x} = 27. 9^{\circ} + 9 0^{\circ} - 6 0^{\circ} = 57. 9^{\circ}$ ; while $δ_{m} = 2 sin^{- 1} (0.75) - 6 0^{\circ} = 37. 2^{\circ}$ . Every achievable deviation lies between these two values.

5. Thin Prism

If $A$ is small (about $1 0^{\circ}$ or less) and the incidence is near normal, all angles are small and $sin θ \approx θ$ . Then $i = n r_{1}$ , $e = n r_{2}$ and

δ = i + e - A = n (r_{1} + r_{2}) - A = (n - 1) A

Figure 7: Thin prism (

A = 1 0^{\circ}

): small angles make

sin θ \approx θ

, so

δ \approx (n - 1) A

. Exact tracing gives

5.0 3^{\circ}

against the formula's

5.0 0^{\circ}

. The deviation hardly depends on

i

as long as

i

is small.

Exam Trick

A thin prism deviates by $(n - 1) A$ whatever the (small) angle of incidence. So a thin prism in front of a mirror, lens or eye simply tilts the beam by a fixed angle: a $4^{\circ}$ prism of $n = 1.5$ always tilts it by $2^{\circ}$ .

Key idea

Thin prism:

δ = (n - 1) A

, independent of small

i

6. Dispersion by a Prism

White light entering a prism emerges as a band of colours, red deviated least and violet most. This splitting is dispersion. It happens because all colours travel at the same speed in vacuum but at slightly different speeds in glass, so $n$ depends on wavelength. Cauchy's relation gives $n (λ) = a + \frac{b}{λ ^{2}}$ , with $a$ , $b$ positive constants of the material.

Figure 8: Dispersion (spread exaggerated:

n

from

1.62

for red to

1.68

for violet, traced exactly). Each colour has its own

n

, so its own deviation; the angle between the red and violet rays is the angular dispersion.

Figure 9: Cauchy's relation

n = a + \frac{b}{λ ^{2}}

for crown glass (

a = 1.5046

b = 4200 nm^{2}

). Shorter wavelengths see a larger

n

, so violet is bent most. The spread

n_{v} - n_{r}

is tiny compared with

n - 1

, which is why prisms deviate much more than they disperse.

Syllabus note: refraction through a prism is in the rationalised NCERT; angular dispersion, dispersive power and prism combinations are no longer there but still appear in JEE Advanced and in older papers.

For a thin prism each colour is deviated by $(n - 1) A$ with its own $n$ :

Angular dispersion: $θ = δ_{v} - δ_{r} = (n_{v} - n_{r}) A$ .
Mean deviation (yellow): $δ_{y} = (n_{y} - 1) A$ , with $n_{y} \approx \frac{n _{v} + n _{r}}{2}$ if not given.
Dispersive power (a property of the material, not of $A$ ): $ω = \frac{θ}{δ _{y}} = \frac{n _{v} - n _{r}}{n _{y} - 1}$ . Flint glass has roughly twice the dispersive power of crown glass.

Figure 10:

δ

i

curves for red (

n_{r} = 1.5132

) and violet (

n_{v} = 1.5309

) light in the

6 0^{\circ}

crown glass prism of Figure 9. Violet is deviated more at every

i

δ_{m} = 39. 9^{\circ}

against

38. 3^{\circ}

, a spread of about

1. 6^{\circ}

. That is larger than the thin-prism value

(n_{v} - n_{r}) A = 1.0 6^{\circ}

, which holds only for small

A

. Violet also has the smaller critical angle, so its curve starts later (

30. 3^{\circ}

against

28. 9^{\circ}

Achromatic combination (deviation without dispersion)

Crown and flint prisms in contact, one reversed, with $(n_{v} - n_{r}) A = (n_{v}^{'} - n_{r}^{'}) A^{'}$ , i.e. $ω δ + ω^{'} δ^{'} = 0$ . Colours recombine; net deviation $δ_{y} (1 - \frac{ω}{ω ^{'}})$ remains.

Direct-vision combination (dispersion without deviation)

Prisms chosen with $(n_{y} - 1) A = (n_{y}^{'} - 1) A^{'}$ , so the mean ray is undeviated. Net dispersion $(n_{y} - 1) A (ω - ω^{'})$ remains; used in direct-vision spectroscopes.

Figure 11: Two thin prisms in contact, one reversed (schematic, angles exaggerated). Left, achromatic prism:

(n_{v} - n_{r}) A = (n_{v}^{'} - n_{r}^{'}) A^{'}

, so the colours recombine but the beam is still deviated. Right, direct-vision prism:

(n_{y} - 1) A = (n_{y}^{'} - 1) A^{'}

, so yellow is undeviated but the colours spread.

Quick Recall: tap to check

Which colour has the smallest critical angle in glass, and which is deviated most?

Violet on both counts: it has the largest

n

Does dispersive power depend on the prism angle?

No.

ω = \frac{n _{v} - n _{r}}{n _{y} - 1}

depends only on the material.

Key idea

Deviation depends on

n - 1

, dispersion on

n_{v} - n_{r}

; two different glasses let you cancel one while keeping the other.

7. Flowchart and Mind Map

Classify the problem first: thin prism, minimum deviation, or a general face-by-face trace.

Figure 12: Problem-solving flowchart. Thin prism, then minimum deviation, then the general face-by-face trace, always checking the second face for total internal reflection.

Figure 13: Prism on one page. Revise from the map, then test yourself on the solved examples.

8. Solved Examples

Solved Example 1

A ray strikes one face of a

6 0^{\circ}

prism (

n = 1.5

) at

6 0^{\circ}

. Find the angle of emergence and the deviation. Which other angle of incidence gives the same deviation?

Solution:

$sin r_{1} = \frac{sin 6 0 ^{\circ}}{1.5} = 0.577$ , $r_{1} = 35. 3^{\circ}$ . $r_{2} = 6 0^{\circ} - 35. 3^{\circ} = 24. 7^{\circ}$ . $sin e = 1.5 sin 24. 7^{\circ} = 0.627$ , $e = 38. 9^{\circ}$ . $δ = 6 0^{\circ} + 38. 9^{\circ} - 6 0^{\circ} = 38. 9^{\circ}$ .

Answer: $e \approx 38. 9^{\circ}$ , $δ \approx 38. 9^{\circ}$ ; incidence at $38. 9^{\circ}$ (the reversed path) gives the same deviation.

Solved Example 2

A ray incident at

6 0^{\circ}

on a

3 0^{\circ}

prism suffers a deviation of

3 0^{\circ}

. Show that it leaves normal to the second face and find

n

Solution:

$δ = i + e - A$ : $3 0^{\circ} = 6 0^{\circ} + e - 3 0^{\circ}$ , so $e = 0$ : the ray emerges normally. Then $r_{2} = 0$ , $r_{1} = A = 3 0^{\circ}$ and $n = \frac{sin 6 0 ^{\circ}}{sin 3 0 ^{\circ}}$ .

Answer: $n = 3 \approx 1.73$ .

Solved Example 3

A ray passes symmetrically through an equilateral prism with an angle of incidence of

4 9^{\circ}

. The refractive index is
(A)

1.33

(B)

1.41

(C)

1.51

(D)

1.73

Solution:

Symmetric passage is minimum deviation: $r = \frac{A}{2} = 3 0^{\circ}$ , so $n = \frac{sin 4 9 ^{\circ}}{sin 3 0 ^{\circ}} = \frac{0.7547}{0.5} = 1.51$ .

Answer: (C).

Solved Example 4

The angle of minimum deviation of an equilateral prism equals its refracting angle. The refractive index of the prism is
(A)

2

(B)

1.5

(C)

3

(D)

2

Solution:

$δ_{m} = A = 6 0^{\circ}$ : $n = \frac{sin \frac{6 0 ^{\circ} + 6 0 ^{\circ}}{2}}{sin 3 0 ^{\circ}} = \frac{sin 6 0 ^{\circ}}{sin 3 0 ^{\circ}} = 3$ .

Answer: (C).

Solved Example 5

6 0^{\circ}

prism has

n = 1.632

. Find its angle of minimum deviation.

Solution:

$sin \frac{6 0 ^{\circ} + δ _{m}}{2} = 1.632 sin 3 0^{\circ} = 0.816$ , so $\frac{6 0 ^{\circ} + δ _{m}}{2} = 54.6 9^{\circ}$ and $δ_{m} = 49.3 7^{\circ}$ .

Answer: $δ_{m} \approx 49. 4^{\circ}$ ( $4 9^{\circ} 2 2^{'}$ ).

Solved Example 6

Find the minimum and maximum deviation produced by a

6 0^{\circ}

prism with

n = 1.5

Solution:

Minimum: $δ_{m} = 2 sin^{- 1} (1.5 sin 3 0^{\circ}) - 6 0^{\circ} = 2 sin^{- 1} (0.75) - 6 0^{\circ} = 37. 2^{\circ}$ .

Maximum (grazing incidence): $r_{1} = C = 41. 8^{\circ}$ , $r_{2} = 18. 2^{\circ}$ , $sin e = 1.5 sin 18. 2^{\circ} = 0.468$ , $e = 27. 9^{\circ}$ , $δ = 9 0^{\circ} + 27. 9^{\circ} - 6 0^{\circ} = 57. 9^{\circ}$ .

Answer: $37. 2^{\circ}$ and $57. 9^{\circ}$ (the ends of Figure 3).

Solved Example 7

For a

6 0^{\circ}

prism with

n = \frac{7}{3}

, find the smallest angle of incidence for which light emerges from the second face, and the maximum deviation.

Solution:

$sin C = \frac{3}{7}$ , $cos C = \frac{4}{7}$ . Grazing emergence: $sin i_{m i n} = n sin (6 0^{\circ} - C) = \frac{7}{3} (\frac{3}{2} \frac{4}{7} - \frac{1}{2} \frac{3}{7}) = 1 - \frac{1}{2} = \frac{1}{2}$ .

Answer: $i_{m i n} = 3 0^{\circ}$ and $δ_{m a x} = 3 0^{\circ} + 9 0^{\circ} - 6 0^{\circ} = 6 0^{\circ}$ .

Solved Example 8

A ray falls normally on one face of a prism of angle

3 0^{\circ}

and

n = 2

. The deviation is
(A)

0^{\circ}

(B)

1 5^{\circ}

(C)

3 0^{\circ}

(D)

4 5^{\circ}

Solution:

$i = 0$ gives $r_{1} = 0$ , $r_{2} = 3 0^{\circ}$ . $sin e = 2 sin 3 0^{\circ} = \frac{1}{2}$ , $e = 4 5^{\circ}$ . $δ = 0 + 4 5^{\circ} - 3 0^{\circ} = 1 5^{\circ}$ .

Answer: (B).

Solved Example 9

What is the largest angle a glass prism (

n = 1.5

) can have if any light at all is to pass through its two refracting faces?

Solution:

Light can emerge only if $A \leq 2 C$ . $C = sin^{- 1} \frac{2}{3} = 41. 8^{\circ}$ .

Answer: $A_{m a x} = 2 C \approx 83. 6^{\circ}$ (only near-grazing rays emerge close to this limit).

Solved Example 10

A thin prism of angle

4^{\circ}

and

n = 1.5

stands in front of a vertical plane mirror. A horizontal ray passes through the prism and hits the mirror. Find the angle of incidence at the mirror.

Solution:

Thin prism deviation: $δ = (1.5 - 1) \times 4^{\circ} = 2^{\circ}$ . The ray reaches the mirror $2^{\circ}$ from the horizontal, i.e. $2^{\circ}$ from the mirror's normal.

Answer: $2^{\circ}$ .

Solved Example 11

The refractive indices of flint glass for red and violet light are

1.613

and

1.632

. Find the angular dispersion produced by a thin flint prism of angle

5^{\circ}

Solution:

$θ = (n_{v} - n_{r}) A = (1.632 - 1.613) \times 5^{\circ}$ .

Answer: $0.09 5^{\circ}$ .

Solved Example 12

Find the dispersive powers of crown glass (

n_{v} = 1.522

n_{r} = 1.514

) and flint glass (

n_{v} = 1.662

n_{r} = 1.644

Solution:

Crown: $n_{y} = 1.518$ , $ω = \frac{0.008}{0.518} = 0.0154$ . Flint: $n_{y} = 1.653$ , $ω^{'} = \frac{0.018}{0.653} = 0.0276$ .

Answer: $0.0154$ and $0.0276$ : flint disperses about $1.8$ times as strongly.

Solved Example 13

A prism of angle

4. 2^{\circ}

(

ω^{'} = 0.045

n^{'} = 1.65

) is to be combined with a prism of

ω = 0.021

n = 1.53

to form an achromatic combination. Find the angle of the second prism and the net deviation.

Solution:

No dispersion: $ω (n - 1) A + ω^{'} (n^{'} - 1) A^{'} = 0$ , so $A = - \frac{0.045 \times 0.65 \times 4. 2 ^{\circ}}{0.021 \times 0.53} = - 11.0 4^{\circ}$ (the minus sign means reversed).

Net deviation $= (0.53) (- 11.0 4^{\circ}) + (0.65) (4. 2^{\circ}) = - 5.8 5^{\circ} + 2.7 3^{\circ}$ .

Answer: $A \approx 11. 0^{\circ}$ , reversed; net deviation $\approx 3.1 2^{\circ}$ , towards the base of the larger prism.

Solved Example 14

A crown prism of angle

6^{\circ}

(

n_{v} = 1.525

n_{r} = 1.515

n_{y} = 1.52

) is combined with a flint prism (

n_{v}^{'} = 1.632

n_{r}^{'} = 1.608

n_{y}^{'} = 1.62

) to give no mean deviation. Find the flint prism angle and the net angular dispersion.

Solution:

No mean deviation: $(n_{y} - 1) A = (n_{y}^{'} - 1) A^{'}$ , so $A^{'} = \frac{0.52 \times 6 ^{\circ}}{0.62} = 5.0 3^{\circ}$ (reversed).

Net dispersion $= (n_{v} - n_{r}) A - (n_{v}^{'} - n_{r}^{'}) A^{'} = 0.010 \times 6^{\circ} - 0.024 \times 5.0 3^{\circ} = 0.06 0^{\circ} - 0.12 1^{\circ}$ .

Answer: $A^{'} \approx 5. 0^{\circ}$ ; net angular dispersion $\approx 0.06 1^{\circ}$ , in the sense of the flint prism.

Practice Questions

A $6 0^{\circ}$ prism has $n = 1.5$ . Find the angle of incidence for minimum deviation and $δ_{m}$ .Answer: $i = sin^{- 1} 0.75 \approx 48. 6^{\circ}$ ; $δ_{m} = 2 i - 6 0^{\circ} \approx 37. 2^{\circ}$
A $6 0^{\circ}$ prism has $n = 2$ . Find $δ_{m}$ .Answer: $3 0^{\circ}$ ( $i = 4 5^{\circ}$ )
Find the deviation produced by a thin prism of angle $5^{\circ}$ and $n = 1.6$ .Answer: $3^{\circ}$
The refractive indices of a glass for red and violet are $1.50$ and $1.60$ . Estimate $n_{y}$ and the dispersive power.Answer: $n_{y} \approx 1.55$ , $ω \approx 0.18$
A crown prism of $5^{\circ}$ ( $n_{v} = 1.523$ , $n_{r} = 1.515$ ) is combined with a flint prism ( $n_{v} = 1.688$ , $n_{r} = 1.650$ ) for dispersion without deviation. Find the flint prism angle.Answer: $A^{'} = \frac{0.519 \times 5 ^{\circ}}{0.669} \approx 3. 9^{\circ}$
The minimum deviation of a $6 0^{\circ}$ prism is $3 8^{\circ}$ . Find $n$ .Answer: $\frac{sin 4 9 ^{\circ}}{sin 3 0 ^{\circ}} \approx 1.51$
For which prism angles of a glass with $C = 4 0^{\circ}$ will every entering ray emerge from the second face?Answer: $A \leq 4 0^{\circ}$

Common Mistakes to Avoid

Watch out

Writing $δ = i - e$ or $δ = i + e + A$ . The deviation is $δ = i + e - A$ .
Using $r = \frac{A}{2}$ when the path is not symmetric. That holds only at minimum deviation.
Applying $δ = (n - 1) A$ to a $6 0^{\circ}$ prism. It is valid only for thin prisms at small incidence.
Forgetting that $i$ and $e$ can be swapped, so most deviations occur at two angles of incidence.
Ignoring total internal reflection at the second face: check $r_{2} < C$ before using Snell's law there.
Using the prism formulas for a glass slab. Parallel faces give no deviation, only a lateral shift.
Treating dispersive power as depending on the prism angle. $ω$ is a property of the glass.
In achromatic combinations, forgetting that one prism must be reversed (its angle is negative in the formula).

Frequently Asked Questions

What is the angle of deviation of a prism?

The angle of deviation is the angle between the direction of the incident ray and the emergent ray. For a prism of angle A, it equals i plus e minus A, where i is the angle of incidence and e the angle of emergence. The ray always bends towards the base of the prism.

What is the condition for minimum deviation in a prism?

Deviation is minimum when the ray passes symmetrically: the angle of incidence equals the angle of emergence and the angles inside the prism are both A over 2. In an isosceles prism the ray then travels parallel to the base. There is only one angle of incidence that gives minimum deviation.

How is the refractive index of a prism found from minimum deviation?

At minimum deviation r equals A over 2 and i equals A plus delta m over 2. Snell's law then gives n equals sin of half of A plus delta m divided by sin of half of A. Measuring delta m on a spectrometer gives the refractive index of the glass accurately.

Why does a thin prism deviate light by (n - 1)A?

For a thin prism and small angle of incidence all angles are small, so sines can be replaced by the angles themselves. Then i equals n r1 and e equals n r2, and the deviation i plus e minus A becomes n times A minus A, that is (n - 1)A, independent of the angle of incidence.

Why does a prism split white light into colours?

The refractive index of glass depends slightly on wavelength: it is larger for violet than for red. Each colour is therefore deviated by a different amount, violet the most and red the least, and white light spreads into a spectrum. This is called dispersion.

What is dispersive power?

Dispersive power is the ratio of angular dispersion to mean deviation, equal to n violet minus n red divided by n yellow minus 1. It depends only on the material, not on the prism angle. Flint glass has roughly twice the dispersive power of crown glass.

Is the prism important for JEE Main and JEE Advanced?

Yes. JEE Main regularly asks minimum deviation, the refractive index formula, thin prism deviation and the deviation against incidence graph. JEE Advanced adds grazing incidence and emergence, conditions for no emergence using the critical angle, and achromatic or direct vision prism combinations.

Which prism questions come in NEET?

NEET mostly asks the prism formula at minimum deviation, thin prism deviation (n - 1)A, the relation r1 plus r2 equals A, and conceptual questions on the deviation graph and on why violet light deviates most. Practise finding n from the angle of minimum deviation.

Previous year questions on Prism

19 questions from past papers, each with a step-by-step solution.

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