Refraction through a Prism

Ray Optics and Optical Instruments · CBSE

Theory

What this subtopic is examined on

The geometry: $r_1 + r_2 = A$ and $\delta = i + e - A$ , derived from the quadrilateral and the triangle inside the prism.
The $\delta$ versus $i$ graph and what happens at minimum deviation.
The derivation of the prism formula for the refractive index (a repeated 3- and 5-mark question).
Thin prisms, dispersion, and the effect of immersing the prism in a liquid.
Ray-tracing questions: normal incidence on one face, grazing emergence, total internal reflection inside a prism.

The geometry of a prism

Diagram in words

Draw a triangular prism $ABC$ with the refracting angle $A$ at the apex and $BC$ as the base. A ray $PQ$ strikes face $AB$ at $Q$ at angle of incidence $i$ , refracts into the glass at angle $r_1$ , travels to $R$ on face $AC$ where it strikes at angle $r_2$ , and emerges into air at angle of emergence $e$ along $RS$ .
Draw the normals at $Q$ and $R$ ; they meet inside at $N$ . Produce the incident ray $PQ$ forwards and the emergent ray $SR$ backwards; the angle between these two directions is the angle of deviation $\delta$ .

The two relations

In the quadrilateral $AQNR$ , the angles at $Q$ and $R$ are right angles, so the remaining two sum to $180^\circ$ :

$\angle A + \angle QNR = 180^\circ$

In triangle $QNR$ , $r_1 + r_2 + \angle QNR = 180^\circ$ . Comparing,

$r_1 + r_2 = A$

The total deviation is the sum of the deviations at the two faces, $(i-r_1)$ at the first and $(e-r_2)$ at the second:

$\delta = (i - r_1) + (e - r_2) = (i + e) - (r_1 + r_2)$

$\Rightarrow\quad \delta = i + e - A \qquad\text{or}\qquad i + e = A + \delta$

The deviation curve and minimum deviation

Plot $\delta$ against $i$ : the curve falls from a large value at small $i$ , reaches a single minimum $\delta_m$ , and rises again. Label the axes $i$ (horizontal) and $\delta$ (vertical) and mark $\delta_m$ and the corresponding $i$ .
Any value of $\delta$ except $\delta_m$ corresponds to two angles of incidence, because $i$ and $e$ appear symmetrically in $\delta = i+e-A$ : the path can be traced backwards for the same deviation.
At minimum deviation the ray passes symmetrically through the prism:

$i = e, \qquad r_1 = r_2 = r, \qquad \text{the refracted ray inside is parallel to the base}$

From $r_1+r_2 = A$ : $\;2r = A$ , so $r = \dfrac{A}{2}$ .
From $\delta = i+e-A$ with $i = e$ : $\;\delta_m = 2i - A$ , so $i = \dfrac{A+\delta_m}{2}$ .

The prism formula

Applying Snell's law at the first face at minimum deviation,

$n_{21} = \dfrac{\sin i}{\sin r} = \dfrac{\sin\left(\dfrac{A+\delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}$

$A$ and $\delta_m$ are both measurable with a spectrometer, so this equation is the standard laboratory method of finding the refractive index of a transparent material.
If the prism sits in a medium rather than in air, the formula returns the relative index $n_{prism}/n_{medium}$ .

Thin prisms

For a small refracting angle $A$ , $\delta_m$ is also small, and the sines may be replaced by the angles:

$n_{21} \approx \dfrac{(A+\delta_m)/2}{A/2} \qquad\Rightarrow\qquad \delta_m = (n_{21}-1)A$

Thin prisms therefore deviate light only slightly, and the deviation is proportional to the refracting angle.

Dispersion

Refractive index falls as wavelength rises, so $n_{violet} > n_{red}$ .
Since $\delta_m$ increases with $n$ , violet is deviated most and red least: white light entering a prism is split into its constituent colours. Red therefore has the smallest angle of minimum deviation of the visible colours.
Assertion-and-reason form CBSE has used: " $\delta_m$ for a prism is smaller for red than for blue" is true, and "the refractive index is smaller for red than for blue" is the correct explanation of it.

A prism immersed in a liquid

What matters is the refractive index of the prism relative to its surroundings. In water, $^w n_g = n_g/n_w$ is smaller than $n_g$ , so the deviation is smaller: $\delta_m$ decreases when the prism is immersed in water.
Example: a prism of $n = \sqrt3 \approx 1.73$ in water of $n = 4/3$ has a relative index of $\dfrac{3\sqrt3}{4} \approx 1.30$ , so $\delta_m$ falls.
Similarly, if the ray emerges into a liquid rather than into air, the angle of emergence is smaller: with $n\sin r_2 = n_L\sin e$ , a larger $n_L$ gives a smaller $e$ .

Ray tracing inside a prism

Normal incidence on one face means $r_1 = 0$ , so $r_2 = A$ : the ray meets the second face at an angle of incidence equal to the refracting angle.
Compare that angle with the critical angle $i_c = \sin^{-1}(1/n)$ for the prism material against whatever lies outside. If it exceeds $i_c$ , the ray is totally internally reflected; if not, it refracts out.
Grazing emergence ( $e = 90^\circ$ ) means the angle of incidence inside at the second face is exactly the critical angle, $r_2 = i_c$ ; then $r_1 = A - i_c$ , and Snell's law at the first face gives the required $i$ .
A right-angled isosceles prism with a ray entering normally on a perpendicular face meets the hypotenuse at $45^\circ$ ; total internal reflection there needs $i_c \le 45^\circ$ , i.e. $n \ge \sqrt2 \approx 1.414$ .

Worked patterns CBSE repeats

Equilateral prism with $\delta_m = A = 60^\circ$ : $n = \dfrac{\sin 60^\circ}{\sin 30^\circ} = \sqrt3 \approx 1.732$ . In general when $\delta_m = A$ , $\;n = \dfrac{\sin A}{\sin(A/2)}$ .
Equilateral prism, $i = 45^\circ$ with a symmetric path: $\delta_m = 2(45^\circ) - 60^\circ = 30^\circ$ , $n = \dfrac{\sin 45^\circ}{\sin 30^\circ} = \sqrt2$ , and $r_1 = r_2 = 30^\circ$ . The speed in the prism is $v = c/n = 3\times10^8/1.414 \approx 2.12\times10^8$ m s $^{-1}$ .
Equilateral prism, $i = \dfrac34 A = 45^\circ$ at minimum deviation: same as above, $n = \sqrt2$ , $v \approx 2.12\times10^8$ m s $^{-1}$ .
Equilateral prism of $n = \sqrt2$ with the emergent ray grazing face $AC$ : $\sin r_2 = 1/\sqrt2$ so $r_2 = 45^\circ$ ; and the prism formula with $A = 60^\circ$ gives $\sin\left(30^\circ + \dfrac{\delta_m}{2}\right) = \dfrac{1}{\sqrt2}$ , so $\delta_m = 30^\circ$ .
Equilateral prism with $\delta_m = 30^\circ$ , emergent ray to graze $AC$ : $n = \sqrt2$ , $r_2 = i_c = 45^\circ$ , $r_1 = 60^\circ - 45^\circ = 15^\circ$ , and $\sin i = \sqrt2\sin 15^\circ = \dfrac{\sqrt3-1}{2} \approx 0.366$ , giving $i \approx 21.5^\circ$ .
Prism with $\angle B = 60^\circ$ , $\angle C = 90^\circ$ , so $A = 30^\circ$ , $\mu = 1.5$ , ray entering face $BA$ normally: $r_2 = 30^\circ$ , and $i_c = \sin^{-1}(1/1.5) \approx 42^\circ$ . Since $30^\circ < 42^\circ$ , there is no total internal reflection and the ray emerges from face $AC$ .
Same prism shape with $n = \sqrt3$ and the ray entering face $AC$ normally: it meets $AB$ at $30^\circ$ , and $\sqrt3\sin30^\circ = \sin e$ gives $e = 60^\circ$ . Emerging instead into a liquid of index 1.3, $\sin e' = 0.866/1.3 = 0.666$ , so $e' \approx 41.8^\circ$ — the angle of emergence decreases.
Isosceles right-angled prism, rays incident normally on $AB$ and striking $AC$ at $45^\circ$ . For $n = 1.33$ , $C \approx 48.8^\circ > 45^\circ$ , so the ray refracts out; for $n = 1.45$ , $C \approx 43.6^\circ < 45^\circ$ , so it is totally internally reflected. With $n = 1.39,\ 1.47,\ 1.52$ the critical angles are about $46^\circ$ , $43^\circ$ and $41^\circ$ : the first refracts out, the other two are totally internally reflected.
Equilateral glass prism ( $\mu_g = 3/2$ ) in water ( $\mu_w = 4/3$ ), ray incident normally on $AB$ : the angle at $AC$ is $60^\circ$ , and $\sin\theta_c = \dfrac{4/3}{3/2} = \dfrac89 \approx 0.889$ while $\sin 60^\circ \approx 0.866$ . Since $\sin 60^\circ < \sin\theta_c$ , the angle is below the critical angle and the ray refracts out of $AC$ .
Refracted ray parallel to the base implies the symmetric path, hence $i = e$ , $r_1 = r_2 = A/2$ and $\delta_m = 2i - A$ .

Standard results

$r_1 + r_2 = A$ ; $\;i + e = A + \delta$ .
At minimum deviation: $i = e$ , $r_1 = r_2 = A/2$ , refracted ray parallel to the base, $\delta_m = 2i - A$ .
$n = \dfrac{\sin\left(\dfrac{A+\delta_m}{2}\right)}{\sin\left(\dfrac{A}{2}\right)}$ ; thin prism $\delta_m = (n-1)A$ .
$\delta_m$ is smallest for red, largest for violet; it decreases when the prism is immersed in a denser medium.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
Given below are two statements labelled as Assertion (A) and Reason (R). Assertion (A): The angle of minimum deviation for a prism is smaller for red light than for blue light. Reason (R): The refractive index of the material of the prism is smaller for red light than for blue light.
- (a)
  Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- (b)
  Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- (c)
  Assertion (A) is true, but Reason (R) is false.
- (d)
  Assertion (A) is false but Reason (R) is true.
1 markCBSE Class 12 Physics — Very Short Answer · 2021, U · Section A, 1 markAnswer 1 ↓
2
A ray of light, on passing through an equilateral glass prism, suffers a minimum deviation equal to the angle of the prism. The value of the refractive index of the material of the prism is ____.
1 markCBSE Class 12 Physics — Very Short Answer · 2020 · Section A, 1 markAnswer 2 ↓
3
How does the angle of minimum deviation of a glass prism vary, if the incident violet light is replaced by red light? Give reason.
1 markCBSE Class 12 Physics — Very Short Answer · AI 2017 · Section A, 1 markAnswer 3 ↓
4
A ray PQ, incident normally on the refracting face BA of a prism made of a material of refractive index 1.5, is shown in the figure: the prism has vertices A (top), B (bottom-left, with $\angle B=60°$ marked) and C (bottom-right, with a right angle, $\angle C=90°$ , marked); the ray PQ strikes face BA perpendicularly and travels on towards face AC. Complete the path of the ray through the prism. From which face will the ray emerge? Justify your answer.
2 marksCBSE Class 12 Physics — SA · AI 2016, Ap · Section B, 2 marksAnswer 4 ↓
5
Two monochromatic rays of light, '1' and '2', are incident normally on the face AB of an isosceles right-angled prism ABC ( $\angle A=45°$ , $\angle B=90°$ , $\angle C=45°$ ). The refractive indices of the glass prism for rays 1 and 2 are respectively 1.33 and 1.45. Trace the path of these rays after they enter the prism, justifying your answer with calculations.
3 marksCBSE Class 12 Physics — SA · AI 2014, An · Section C, 3 marksAnswer 5 ↓
6
Figure shows a ray of light PQ passing through a prism ABC (apex angle $A$ ), refracting at Q on face AB and again at R on face AC, with the refracted ray QR parallel to the base BC. If the refracted ray QR is parallel to the base BC, show that (i) $r_1=r_2=A/2$ , (ii) the angle of minimum deviation is $\delta_m=2i-A$ .
3 marksCBSE Class 12 Physics — SA · Foreign 2014 · Section C, 3 marksAnswer 6 ↓
7
(a) Write two necessary conditions for total internal reflection to take place. (b) Two right-angled prisms ABC and DBC are arranged in contact along their common face BC, as shown: prism ABC has $\angle A=45°$ and $\angle B=90°$ (so $\angle C=45°$ ), sitting on top of prism DBC which has $\angle B=60°$ and $\angle D=60°$ (so $\angle C=60°$ ), the two prisms sharing the edge BC. The critical angles for the two prisms (with respect to air) are $41.1°$ and $45°$ respectively. A horizontal ray enters prism ABC through face AB. Trace the path of the ray through the combination.
3 marksCBSE Class 12 Physics — SA · Term II 2021-22, An · Section C, 3 marksAnswer 7 ↓
8
A ray of light passes through an equilateral prism ( $A=60°$ ) of refractive index $\sqrt2$ , entering at M on face AB (refraction angle $r_1$ ) and striking face AC at N (angle $r_2$ ), such that the emergent ray just grazes along face AC. Find: (i) the angle of incidence ( $r_2$ ) at face AC, (ii) the angle of minimum deviation for this prism.
3 marksCBSE Class 12 Physics — SA · Term II 2021-22 · Section C, 3 marksAnswer 8 ↓
9
A ray of light is incident on an equilateral prism ( $A=60°$ ) at an angle of $45°$ and passes symmetrically through it, as shown. Calculate (a) the angle of minimum deviation, (b) the refractive index of the material of the prism, and (c) the angle of refraction at the point of incidence, P.
3 marksCBSE Class 12 Physics — SA · Term II 2021-22 · Section C, 3 marksAnswer 9 ↓
10
(a) Plot a graph showing the variation of the angle of deviation as a function of the angle of incidence, for light passing through a triangular prism. (b) Derive the relation for the refractive index of the material of the prism in terms of the angle of minimum deviation and the angle of the prism.
5 marksCBSE Class 12 Physics — Long Answer · AI 2019 · Section E, 5 marksAnswer 10 ↓
11
A ray of light falls normally on face AB of an equilateral glass prism ( $\mu_{glass}=3/2$ ) placed in water ( $\mu_{water}=4/3$ ), and travels on to strike face AC. Will this ray suffer total internal reflection on striking face AC? Justify your answer.
3 marksCBSE Class 12 Physics — SA · 2018, An · Section C, 3 marksAnswer 11 ↓
12
(a) A ray of light incident on face AB of an equilateral glass prism shows a minimum deviation of $30°$ . Calculate the speed of light through the prism. (b) Find the angle of incidence at face AB so that the emergent ray grazes along face AC (i.e. so that the angle of emergence is $90°$ ).
3 marksCBSE Class 12 Physics — SA · Delhi 2017 · Section C, 3 marksAnswer 12 ↓
13
Three monochromatic rays 1, 2 and 3 fall normally on face AB of an isosceles right-angled prism ABC ( $\angle A=45°$ , $\angle B=90°$ , $\angle C=45°$ ) and strike the hypotenuse AC. The refractive indices of the prism material for rays 1, 2 and 3 are respectively 1.39, 1.47 and 1.52. Find which of these rays get totally internally reflected and which get only refracted from AC. Trace the paths of the rays, justifying your answer with the necessary calculations.
5 marksCBSE Class 12 Physics — Long Answer · Foreign 2016, An · Section E, 5 marksAnswer 13 ↓
14
(i) Explain the working principle of an optical fibre with the help of a diagram. Mention one use of a light pipe. (ii) A ray of light is incident at an angle of $60°$ on one face of a prism of apex angle $A=60°$ , and the ray passes symmetrically through the prism. Find the angle of minimum deviation ( $\delta_m$ ) and the refractive index of the material of the prism. If the prism is immersed in water, how will $\delta_m$ be affected? Justify your answer.
5 marksCBSE Class 12 Physics — Long Answer · 2023 · Section E, 5 marksAnswer 14 ↓
15
A ray of light is incident normally on face AC of a glass prism ABC of refractive index $\sqrt3$ (with $\angle B=60°$ and $\angle C=90°$ , so $\angle A=30°$ ), and after travelling through the prism, strikes face AB and emerges into air. Calculate the angle of emergence. How will the angle of emergence change qualitatively if the ray emerges from the prism into a liquid of refractive index 1.3 instead of air?
5 marksCBSE Class 12 Physics — Long Answer · 2020, Ev · Section E, 5 marksAnswer 15 ↓
16
A ray of light, passing from air through an equilateral glass prism, undergoes minimum deviation when the angle of incidence is $\dfrac34$ of the angle of the prism. Calculate the speed of light in the prism.
2 marksCBSE Class 12 Physics — SA · AI 2017, Ap · Section B, 2 marksAnswer 16 ↓
17
(a) Plot a graph showing the variation of the angle of deviation as a function of the angle of incidence for light passing through a prism. Derive an expression for the refractive index of the prism in terms of the angle of minimum deviation and the angle of the prism. (b) A ray of light, incident normally on one face of a right isosceles prism, is totally internally reflected, as shown. What must be the minimum value of the refractive index of the glass? Give relevant calculations.
5 marksCBSE Class 12 Physics — Long Answer · Section E, 5 marksAnswer 17 ↓
18
(a) A ray PQ of light is incident on face AB of a glass prism ABC and emerges out of face AC. Trace the path of the ray and show that $\angle i+\angle e=\angle A+\angle\delta$ , where $\delta$ and $e$ denote the angle of deviation and the angle of emergence respectively. Plot a graph showing the variation of the angle of deviation as a function of the angle of incidence. State the condition under which $\angle\delta$ is minimum. (b) Find the relation between the refractive index ( $\mu$ ) of the glass prism and $\angle A$ , for the case when the angle of the prism $A$ is equal to the angle of minimum deviation $\delta_m$ . Hence obtain the value of the refractive index for a prism angle $A=60°$ .
5 marksCBSE Class 12 Physics — Long Answer · AI 2015, Ev · Section E, 5 marksAnswer 18 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

19
For a glass prism, the angle of minimum deviation will be smallest for the light of colour
- (a)
  red
- (b)
  blue
- (c)
  yellow
- (d)
  green
1 markCBSE Class 12 Physics — Multiple Choice · 2020 · Section A, 1 markAnswer 19 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.Answer: (A)↑ back to the question
Since $\delta_m=(\mu-1)A$ , a smaller refractive index gives a smaller angle of minimum deviation. Red light has a smaller refractive index than blue light in the (normally dispersive) glass of the prism, so it also has a smaller angle of minimum deviation. Both statements are true, and R correctly explains A. Option (a).
2.↑ back to the question
For an equilateral prism, $A=60°$ . Given $\delta_m=A=60°$ . $\mu=\dfrac{\sin\left(\dfrac{A+\delta_m}{2}\right)}{\sin\left(\dfrac A2\right)}=\dfrac{\sin60°}{\sin30°}=\dfrac{\sqrt3/2}{1/2}=\sqrt3\approx1.732$
3.↑ back to the question
$\lambda_{red}>\lambda_{violet}$ , and since refractive index decreases with increasing wavelength (normal dispersion), $\mu_{red}<\mu_{violet}$ . Since $\delta_m=(\mu-1)A$ , a smaller $\mu$ gives a smaller $\delta_m$ : so the angle of minimum deviation DECREASES when violet light is replaced by red light.
4.↑ back to the question
Since $\angle B=60°$ and $\angle C=90°$ , the apex angle at $A$ (between faces AB and AC) is $180°-60°-90°=30°$ . The ray enters face BA normally, so $r_1=0°$ there. Using the prism relation for the two refracting faces AB and AC meeting at vertex $A$ : $\;A=r_1+r_2\;\Rightarrow\;r_2=A-r_1=30°-0°=30°$ — i.e. the ray strikes face AC at an angle of incidence of $30°$ (measured from the normal to AC). Critical angle of the prism material: $\;\sin i_c=\dfrac1\mu=\dfrac1{1.5}=0.667\;\Rightarrow\;i_c\approx42°$ Since the angle of incidence at AC ( $30°$ ) is less than the critical angle ( $42°$ ), the ray does NOT undergo total internal reflection at AC. It refracts (bending away from the normal) and emerges from face AC.
5.↑ back to the question
Since both rays enter face AB normally, they travel undeviated inside the glass until they strike the hypotenuse AC. By the $45°$ - $45°$ - $90°$ geometry of the prism, each ray strikes face AC at an angle of incidence of $45°$ (measured from the normal to AC). Critical angle for ray 1: $\;\sin C_1=\dfrac1{1.33}=0.752\;\Rightarrow\;C_1\approx48.8°$ Critical angle for ray 2: $\;\sin C_2=\dfrac1{1.45}=0.690\;\Rightarrow\;C_2\approx43.6°$ Ray 1: angle of incidence at AC ( $45°$ ) is LESS than its critical angle ( $48.8°$ ), so ray 1 refracts and emerges out through face AC. Ray 2: angle of incidence at AC ( $45°$ ) is GREATER than its critical angle ( $43.6°$ ), so ray 2 undergoes total internal reflection at face AC and is reflected back into the prism (emerging eventually through face BC).
6.↑ back to the question
(i) Since QR is parallel to the base BC, by geometry the two base angles of the triangle formed by QR and the normals are equal, which forces the ray to be travelling symmetrically inside the prism: the angle of incidence at the first face equals the angle of emergence at the second face, i.e. $i=e$ , and correspondingly $r_1=r_2$ . Since $r_1+r_2=A$ (the prism relation) and $r_1=r_2$ : $\;2r_1=A\;\Rightarrow\;r_1=r_2=\dfrac A2$ (ii) The general relation for a prism is $A+\delta=i+e$ , where $\delta$ is the angle of deviation. When the ray travels symmetrically ( $i=e$ ), the deviation is minimum, i.e. $\delta=\delta_m$ : $\;A+\delta_m=i+i=2i$ $\Rightarrow\;\delta_m=2i-A$
7.↑ back to the question
(a) Two necessary conditions for total internal reflection: (i) light must be travelling from an optically denser medium towards an optically rarer medium; (ii) the angle of incidence in the denser medium must be greater than the critical angle for that pair of media, i.e. $\sin i>\sin C=\dfrac{n_2}{n_1}$ (for $n_1>n_2$ ). (b) The horizontal ray entering face AB of the upper prism (ABC) travels on and strikes the slant face AC. Since AC makes $45°$ with the horizontal, the ray meets AC at an angle of incidence of $90°-45°=45°$ from the normal to AC. This ( $45°$ ) is greater than the critical angle of prism ABC ( $41.1°$ ), so the ray undergoes total internal reflection at face AC, turning through $90°$ so that it now travels vertically downward, parallel to AB and perpendicular to BC. This vertically-travelling ray strikes the shared interface BC normally (angle of incidence $=0°$ ), so it passes straight through undeviated into the lower prism DBC. Inside prism DBC, the ray (now travelling vertically) strikes the slant face CD. Since CD makes $60°$ with BC, the ray meets CD at an angle of incidence of $90°-(90°-60°)=60°$ from the normal to CD. This ( $60°$ ) is greater than the critical angle of prism DBC ( $45°$ ), so the ray again undergoes total internal reflection at CD, this time emerging perpendicular to CD travelling horizontally, and finally exits face BD undeviated (striking it normally, angle of incidence $=0°$ ).
8.↑ back to the question
(i) The emergent ray grazing along face AC means the angle of emergence there is $e=90°$ . Applying Snell's law at face AC (glass to air): $\;\mu\sin r_2=\sin90°=1\;\Rightarrow\;\sin r_2=\dfrac1\mu=\dfrac1{\sqrt2}\;\Rightarrow\;r_2=45°$ (ii) For the equilateral prism, $\;\mu=\dfrac{\sin\left(\dfrac{A+\delta_m}2\right)}{\sin\left(\dfrac A2\right)}\;\Rightarrow\;\sqrt2=\dfrac{\sin\left(30°+\dfrac{\delta_m}2\right)}{\sin30°}$ $\Rightarrow\;\sin\left(30°+\dfrac{\delta_m}2\right)=\sqrt2\times\dfrac12=\dfrac1{\sqrt2}\;\Rightarrow\;30°+\dfrac{\delta_m}2=45°\;\Rightarrow\;\delta_m=30°$
9.↑ back to the question
(a) Since the ray passes symmetrically, $i=45°$ corresponds to minimum deviation ( $i=e$ ): $\;\delta_m=2i-A=2(45°)-60°=30°$ (b) $\;\mu=\dfrac{\sin\left(\dfrac{A+\delta_m}2\right)}{\sin\left(\dfrac A2\right)}=\dfrac{\sin45°}{\sin30°}=\dfrac{1/\sqrt2}{1/2}=\sqrt2\approx1.414$ (c) At the symmetric path, $r_1=r_2=A/2=30°$ ; checking with Snell's law at P: $\;\dfrac{\sin i}{\sin r}=\mu\;\Rightarrow\;\sin r=\dfrac{\sin45°}{\sqrt2}=\dfrac{1/\sqrt2}{\sqrt2}=\dfrac12\;\Rightarrow\;r=30°$ — confirming the angle of refraction at P is $30°$ .
10.↑ back to the question
(a) As the angle of incidence $i$ is increased from a small value, the angle of deviation $\delta$ first decreases, reaches a single minimum value $\delta_m$ (at some particular $i$ , where the ray passes symmetrically through the prism, $i=e$ ), and then increases again. The $\delta$ -versus- $i$ graph is thus a smooth curve, concave-up, with a single minimum — it is NOT symmetric about the minimum in general shape but does have $i=e$ at that minimum. (b) For a prism of apex angle $A$ : $\;A=r_1+r_2\qquad\text{...(i)}$ , and the total deviation $\;\delta=(i-r_1)+(e-r_2)=(i+e)-(r_1+r_2)=(i+e)-A\qquad\text{...(ii)}$ At minimum deviation, $\delta=\delta_m$ , $i=e$ , and correspondingly $r_1=r_2=r$ (by symmetry). From (i): $\;r=\dfrac A2$ From (ii): $\;\delta_m=2i-A\;\Rightarrow\;i=\dfrac{A+\delta_m}2$ The refractive index of the prism material is then: $\;\mu=\dfrac{\sin i}{\sin r}=\dfrac{\sin\left(\dfrac{A+\delta_m}2\right)}{\sin\left(\dfrac A2\right)}$
11.↑ back to the question
Since the ray enters face AB normally, it travels undeviated to face AC. For an equilateral prism ( $A=60°$ ), the angle of incidence at face AC works out to $60°$ (using $A=r_1+r_2$ with $r_1=0°$ , so $r_2=A=60°$ ). Critical angle for light going from the glass prism into the surrounding water: $\;\sin\theta_c=\dfrac{\mu_{water}}{\mu_{glass}}=\dfrac{4/3}{3/2}=\dfrac89\approx0.889$ Comparing with the angle of incidence at AC: $\;\sin60°=\dfrac{\sqrt3}2\approx0.866$ Since $\sin60°\,(0.866)<\sin\theta_c\,(0.889)$ , the angle of incidence ( $60°$ ) is LESS than the critical angle. So the ray will NOT undergo total internal reflection — it refracts and emerges from face AC.
12.↑ back to the question
(a) $A=60°$ , $\delta_m=30°$ : $\;\mu=\dfrac{\sin\left(\dfrac{60°+30°}2\right)}{\sin30°}=\dfrac{\sin45°}{\sin30°}=\dfrac{1/\sqrt2}{1/2}=\sqrt2\approx1.414$ Speed of light in the prism: $\;v=\dfrac c\mu=\dfrac{3\times10^8}{1.414}\approx2.12\times10^8$ m/s (b) For the emergent ray to graze along AC, the angle of emergence there must be $90°$ , so the angle of incidence at AC (from inside) equals the critical angle: $\;\sin i_c=\dfrac1\mu=\dfrac1{\sqrt2}\;\Rightarrow\;i_c=r_2=45°$ Using $A=r_1+r_2$ : $\;60°=r_1+45°\;\Rightarrow\;r_1=15°$ Applying Snell's law at face AB: $\;\dfrac{\sin i}{\sin r_1}=\mu\;\Rightarrow\;\sin i=\sqrt2\sin15°=\sqrt2\times\dfrac{\sqrt6-\sqrt2}4=\dfrac{\sqrt3-1}2\approx0.366$ $\Rightarrow\;i=\sin^{-1}\!\left(\dfrac{\sqrt3-1}2\right)\approx21.5°$
13.↑ back to the question
Since all three rays enter face AB normally, they travel undeviated to the hypotenuse AC, striking it at an angle of incidence of $45°$ (by the isosceles right-angle geometry of the prism). Critical angle for ray 1: $\;\sin C_1=\dfrac1{1.39}=0.719\;\Rightarrow\;C_1\approx46°$ Critical angle for ray 2: $\;\sin C_2=\dfrac1{1.47}=0.680\;\Rightarrow\;C_2\approx43°$ Critical angle for ray 3: $\;\sin C_3=\dfrac1{1.52}=0.658\;\Rightarrow\;C_3\approx41°$ Ray 1: angle of incidence at AC ( $45°$ ) is LESS than its critical angle ( $46°$ ), so ray 1 refracts and emerges from face AC. Ray 2: angle of incidence at AC ( $45°$ ) is GREATER than its critical angle ( $43°$ ), so ray 2 undergoes total internal reflection at AC. Ray 3: angle of incidence at AC ( $45°$ ) is GREATER than its critical angle ( $41°$ ), so ray 3 also undergoes total internal reflection at AC.
14.↑ back to the question
(i) An optical fibre consists of a core of high refractive index glass/quartz (typically $\mu\approx1.7$ ) surrounded by a cladding of slightly lower refractive index (typically $\mu\approx1.5$ ). Light launched into the core at a suitable angle repeatedly strikes the core-cladding interface at an angle greater than the critical angle and undergoes total internal reflection over and over, so it travels down the length of the fibre with very little loss even if the fibre is bent. Diagram: a long thin cylindrical core, sheathed by the cladding layer, with a ray shown entering at one end and zig-zagging along the fibre via repeated total internal reflections at the core-cladding boundary, until it emerges at the far end. Use of a light pipe (bundle of optical fibres): used to transmit images, e.g. in medical endoscopy to see inside the human body, and in decorative 'light pipes' — a bundle of fibres, each carrying light from a corresponding small area of an image, reproduces the full image at the other end. (ii) Since $i=60°$ and the ray passes symmetrically ( $i=e$ ): $\;\delta_m=2i-A=2(60°)-60°=60°$ $\;\mu=\dfrac{\sin\left(\dfrac{A+\delta_m}2\right)}{\sin\left(\dfrac A2\right)}=\dfrac{\sin60°}{\sin30°}=\dfrac{\sqrt3/2}{1/2}=\sqrt3\approx1.732$ If the prism is immersed in water ( $\mu_{water}=4/3$ ), the refractive index of the prism material relative to water decreases: $\;^w\mu_g=\dfrac{\sqrt3}{4/3}=\dfrac{3\sqrt3}4\approx1.30$ , which is smaller than $\sqrt3\approx1.73$ . Since $\delta_m=(\mu-1)A$ (approximately, for small $A$ ) or more generally increases with $\mu$ , a smaller relative refractive index means the angle of minimum deviation will DECREASE when the prism is immersed in water.
15.↑ back to the question
Since the ray enters face AC normally, it travels undeviated inside the prism up to face AB. Using $A=r_1+r_2$ with $r_1=0°$ (normal incidence at AC) and apex angle $A=30°$ : the ray strikes face AB at an angle of $30°$ from the normal to AB. Applying Snell's law at face AB, going from the prism (denser) into air (rarer): $\;\mu\sin30°=\sin e\;\Rightarrow\;\sqrt3\times\dfrac12=\sin e\;\Rightarrow\;\sin e=\dfrac{\sqrt3}2\;\Rightarrow\;e=60°$ If the ray instead emerges into a liquid of refractive index 1.3: $\;\mu\sin30°=\mu_L\sin e'\;\Rightarrow\;\sqrt3\times\dfrac12=1.3\sin e'\;\Rightarrow\;\sin e'=\dfrac{0.866}{1.3}\approx0.666\;\Rightarrow\;e'\approx41.8°$ Since $e'\,(\approx41.8°)<e\,(60°)$ , the angle of emergence DECREASES when the ray emerges into the liquid instead of into air.
16.↑ back to the question
$A=60°$ , and $i=\dfrac34A=\dfrac34(60°)=45°$ . At minimum deviation, $i=e$ , so: $\;\delta_m=2i-A=2(45°)-60°=30°$ $\;\mu=\dfrac{\sin\left(\dfrac{A+\delta_m}2\right)}{\sin\left(\dfrac A2\right)}=\dfrac{\sin45°}{\sin30°}=\dfrac{0.7071}{0.5}=1.414\approx\sqrt2$ Speed of light in the prism: $\;v=\dfrac c\mu=\dfrac{3\times10^8}{1.414}\approx2.12\times10^8$ m/s
17.↑ back to the question
(a) Same graph and derivation as in Q61 above: the $\delta$ - $i$ curve has a single minimum at $i=e$ , and $\;\mu=\dfrac{\sin\left(\dfrac{A+\delta_m}2\right)}{\sin\left(\dfrac A2\right)}$ . (b) The ray enters normally on one perpendicular face (undeviated) and strikes the hypotenuse of the isosceles right-angled prism at an angle of incidence of $45°$ (from the $45°$ - $45°$ - $90°$ geometry). For total internal reflection to just occur here, the critical angle of the glass must be at most $45°$ : $\;\sin C\le\sin45°\;\Rightarrow\;\dfrac1\mu\le\dfrac1{\sqrt2}\;\Rightarrow\;\mu\ge\sqrt2$ So the minimum refractive index of the glass required is $\mu_{min}=\sqrt2\approx1.414$ .
18.↑ back to the question
(a) At face AB, the deviation produced is $\delta_1=i-r_1$ . At face AC, the deviation produced is $\delta_2=e-r_2$ . The total deviation is $\;\delta=\delta_1+\delta_2=(i-r_1)+(e-r_2)=(i+e)-(r_1+r_2)$ Since $r_1+r_2=A$ : $\;\delta=(i+e)-A\;\Rightarrow\;i+e=A+\delta$ Graph: same $\delta$ -versus- $i$ curve as in Q61/Q68 — decreasing then increasing, with a single minimum. The deviation $\delta$ is minimum when the ray traverses the prism symmetrically, i.e. when $i=e$ (equivalently $r_1=r_2$ ). (b) When $A=\delta_m$ : from $\delta_m=2i-A$ , we get $\;A=2i-A\;\Rightarrow\;i=A$ , and (since $r_1=r_2=A/2$ at minimum deviation) $\;r=\dfrac A2$ $\;\mu=\dfrac{\sin i}{\sin r}=\dfrac{\sin A}{\sin(A/2)}$ For $A=60°$ : $\;\mu=\dfrac{\sin60°}{\sin30°}=\dfrac{\sqrt3/2}{1/2}=\sqrt3\approx1.732$
19.Answer: (A)↑ back to the question
The angle of minimum deviation $\delta_m=(\mu-1)A$ is smallest when $\mu$ is smallest. Since red light has the longest wavelength, it has the smallest refractive index among visible colours in a normally dispersive medium like glass. Hence red light has the smallest angle of minimum deviation. Option (a).