Chapter 1 · Light – Reflection and Refraction

1.3 Refraction of Light and the Refractive Index

Light changes speed when it passes from one material into another, and so it changes direction. One number for each material, the refractive index $n=\frac cv$, and Snell's law $\frac{\sin i}{\sin r}=n_{21}$ predict exactly how much it bends.

Stand a pencil in a glass of water and look from the side: it seems to snap at the surface. A swimming pool looks much shallower than it really is, which is why people misjudge its depth when they jump in. Put a coin in an empty cup, move your head until the rim just hides it, then pour in water without moving — the coin comes back into view. In each case light has crossed a boundary between two transparent materials and changed direction there.

In this section we find out why that happens, state the two laws that light obeys at every such boundary, follow a ray straight through a glass block, and attach to every transparent material one number — its refractive index — that tells us how strongly it bends light. These ideas are exactly what you need for lenses in Section 1.4, for the eye and the prism in Chapter 2, and for the optical fibres that carry the internet under the oceans.

Refraction: Light Bends at a Boundary

Light travels at about $3.0\times10^{8}\ \text{m/s}$ in a vacuum (and almost as fast in air), but more slowly in water and more slowly still in glass. When a beam meets the boundary at a slant, one edge of the beam enters the slower material first and is slowed first, while the other edge is still racing through the faster one. The beam swings round, just as a line of marching students wheels round when the students at one end step from the road onto sand and slow down before the others.

📘 Definition 1.5 — Refraction and the angle of refraction

Refraction is the change in the direction of light when it passes obliquely from one transparent medium into another, caused by the change in the speed of light.

At the point where a ray meets the boundary, draw the normal — the line at $90^\circ$ to the boundary. The angle of incidence $i$ is the angle between the incident ray and the normal; the angle of refraction $r$ is the angle between the refracted ray and the normal. A medium in which light travels more slowly is called optically denser; one in which it travels faster is optically rarer.

Experiments with many pairs of materials always give the same two behaviours:

This explains the pool and the coin. Light from the bottom of the pool bends away from the normal as it leaves the water; your eye traces the ray back in a straight line and so "sees" the bottom higher than it is. The coin's light, bent at the water surface, now clears the rim of the cup and reaches your eye.

📐 Law 1.5 — The laws of refraction

When light passes from one transparent medium (medium 1) into another (medium 2):

  1. The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
  2. For light of a given colour and a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant:
$$\frac{\sin i}{\sin r}=\text{constant}=n_{21}.$$

The second law is called Snell's law of refraction. The constant $n_{21}$ is the refractive index of medium 2 relative to medium 1. If $n_{21}\gt1$, medium 2 is optically denser and the ray bends towards the normal. (For $i=0^\circ$ both sines are zero and the ray goes straight through.)

Interactive 1 — The Refraction Lab

Choose the two media and move the angle of incidence. The refracted ray is calculated from Snell's law. The orange pulses of light travel along the ray at a speed proportional to $\tfrac cn$: watch them bunch up as they enter a slower medium. The faint grey line shows where the ray would have gone without bending, and a weak reflected ray is drawn too. Press Play to sweep the angle.

40
$i$—
$r$—
$\sin i/\sin r$—
$n_{21}=n_2/n_1$—
speed in medium 2—

Refraction through a Rectangular Glass Slab

Lay a rectangular glass block on paper, stick two pins along a slanting line on one side, and look through the block from the other side for two more pins that line up with their images. Joining the pin marks shows the path of the light: it bends towards the normal as it enters the glass at the first face and away from the normal by the same amount as it leaves at the opposite face.

🧭 What the glass slab shows
  • The faces are parallel, so the angle at which the ray meets the second face inside the glass equals the angle of refraction $r$ at the first face.
  • Snell's law is used twice, in opposite directions: air → glass gives $\sin i=n\sin r$, glass → air gives $n\sin r=\sin e$. Hence the angle of emergence $e$ equals $i$: the emergent ray is parallel to the incident ray.
  • The emergent ray is, however, shifted sideways. This perpendicular distance between the incident ray (produced) and the emergent ray is the lateral shift (lateral displacement). It is larger for a thicker slab, a larger angle of incidence and a material of higher refractive index; it is zero at normal incidence.
✏️ Example 1 — Into a glass slab and out again
A ray of light in air strikes the top face of a glass slab ($n=1.50$) at an angle of incidence of $40^\circ$. Find the angle of refraction inside the glass and the angle at which the ray leaves the bottom face.
First face Air → glass: $$\sin r=\frac{\sin 40^\circ}{1.50}=\frac{0.6428}{1.50}=0.4285,\qquad r=25.4^\circ.$$
Inside The faces are parallel, so the ray meets the bottom face at $25.4^\circ$ to the normal there.
Second face Glass → air: $\sin e=1.50\times\sin 25.4^\circ=1.50\times0.4285=0.6428$, so $e=40^\circ$.
Meaning The ray bends towards the normal on entering ($25.4^\circ\lt40^\circ$) and away from it on leaving, and emerges at $40^\circ$ — parallel to the incident ray, but displaced sideways. Check it in Interactive 2.
Interactive 2 — Through a Glass Slab

A ray crosses a slab with parallel faces. Both refractions are computed from Snell's law. The dashed line is the incident ray produced straight on; the red bar is the lateral shift between it and the emergent ray. Change the angle, the thickness and the material and watch the emergent ray stay parallel to the incident ray while the shift changes. Play sweeps the angle of incidence.

455
$i$—
$r$ inside—
$e$ (emergence)—
lateral shift—

The Refractive Index

How strongly a material bends light depends on how much it slows light down. So we measure every transparent material against the fastest possible case — light in a vacuum.

📘 Definition 1.6 — Absolute and relative refractive index

The absolute refractive index $n$ of a medium is the ratio of the speed of light in a vacuum, $c$, to the speed of light $v$ in that medium:

$$n=\frac cv .$$

The refractive index of medium 2 relative to medium 1 is the ratio of the speed of light in medium 1 to that in medium 2:

$$n_{21}=\frac{v_1}{v_2}=\frac{n_2}{n_1}.$$

Refractive indices are ratios of two speeds, so they have no unit. Since nothing travels faster than light in a vacuum, every absolute refractive index is at least $1$ (air: $1.0003$, taken as $1.00$). Note that $n_{12}=\dfrac{1}{n_{21}}$.

Material$n$Material$n$
Air1.0003Crown glass1.52
Ice1.31Rock salt1.54
Water1.33Carbon disulphide1.63
Alcohol (ethanol)1.36Dense flint glass1.65
Kerosene1.44Ruby1.71
Fused quartz1.46Sapphire1.77
Turpentine oil1.47Diamond2.42

Optical density is not mass density. Kerosene floats on water, so it has the smaller mass density, yet its refractive index ($1.44$) is larger than that of water ($1.33$): kerosene is the optically denser of the two. Light is slowest in diamond, which is why a cut diamond bends and traps light so dramatically.

✏️ Example 2 — How fast is light in glass and in diamond?
Find the speed of light in glass ($n=1.50$) and in diamond ($n=2.42$). How many times faster does light travel in the glass? Take $c=3.0\times10^{8}\ \text{m/s}$.
Formula $n=\dfrac cv$, so $v=\dfrac cn$.
Glass $v=\dfrac{3.0\times10^{8}}{1.50}=2.0\times10^{8}\ \text{m/s}$.
Diamond $v=\dfrac{3.0\times10^{8}}{2.42}=1.24\times10^{8}\ \text{m/s}\approx1.2\times10^{8}\ \text{m/s}$.
Ratio $\dfrac{v_\text{glass}}{v_\text{diamond}}=\dfrac{n_\text{diamond}}{n_\text{glass}}=\dfrac{2.42}{1.50}=1.61$: light is about $1.6$ times faster in the glass. Notice the speeds are in the inverse ratio of the refractive indices.
✏️ Example 3 — Relative refractive index from two speeds
Light travels at $2.0\times10^{8}\ \text{m/s}$ in medium A and at $2.5\times10^{8}\ \text{m/s}$ in medium B. (a) Find the absolute refractive index of each. (b) Find the refractive index of A relative to B. (c) Which way does a ray bend when it passes obliquely from B into A?
(a) $n_A=\dfrac{3.0\times10^{8}}{2.0\times10^{8}}=1.5$ and $n_B=\dfrac{3.0\times10^{8}}{2.5\times10^{8}}=1.2$.
(b) $n_{AB}=\dfrac{v_B}{v_A}=\dfrac{2.5\times10^{8}}{2.0\times10^{8}}=1.25$. Check: $\dfrac{n_A}{n_B}=\dfrac{1.5}{1.2}=1.25$. ✓
(c) Light is slower in A, so A is optically denser. Going from B into A, the ray bends towards the normal.

Calculating with Snell's Law

Combining Snell's law with $n_{21}=\dfrac{n_2}{n_1}$ gives a form that is symmetric and easy to remember: $n_1\sin i=n_2\sin r$. The product $n\sin\theta$ is the same on both sides of every boundary.

🧭 Problem-solving strategy — refraction numericals
  1. Sketch the boundary and draw the normal at the point of incidence. Mark $i$ and $r$ from the normal, never from the surface.
  2. Identify medium 1 (where the light comes from) and medium 2 (where it goes).
  3. Write $n_1\sin i=n_2\sin r$ and solve for the unknown. Use your calculator in degree mode.
  4. Check the direction: into a denser medium $r\lt i$; into a rarer medium $r\gt i$.
✏️ Example 4 — From air into water
A ray of light in air meets a water surface ($n=1.33$) at an angle of incidence of $45^\circ$. Find the angle of refraction.
Media Medium 1 is air ($n_1=1.00$), medium 2 is water ($n_2=1.33$).
Snell $$\sin r=\frac{n_1\sin i}{n_2}=\frac{1.00\times0.7071}{1.33}=0.5317,\qquad r=32.1^\circ.$$
Check $32.1^\circ\lt45^\circ$: the ray bends towards the normal, as it must on entering a denser medium. ✓
✏️ Example 5 — Leaving the water
Light from a coin at the bottom of a cup of water meets the water surface at $30^\circ$ to the normal. At what angle does it travel in the air? Use this to explain why the coin seems raised.
Media Now medium 1 is water ($n_1=1.33$) and medium 2 is air ($n_2=1.00$).
Snell $$\sin r=\frac{1.33\times\sin30^\circ}{1.00}=1.33\times0.500=0.665,\qquad r=41.7^\circ.$$
Meaning The ray bends away from the normal ($41.7^\circ\gt30^\circ$). Your eye traces the emerging ray straight back. That line is more slanted ($41.7^\circ$ from the vertical instead of $30^\circ$), so it crosses the vertical line through the coin at a point above the coin: the coin — and the whole bottom of the cup — appears raised.
✏️ Example 6 — Measuring a refractive index in the laboratory
In a glass-slab experiment a student measures an angle of incidence of $60^\circ$ and an angle of refraction of $35^\circ$. Find the refractive index of the glass and the speed of light in it.
Snell $$n=\frac{\sin 60^\circ}{\sin 35^\circ}=\frac{0.8660}{0.5736}=1.51.$$
Speed $v=\dfrac cn=\dfrac{3.0\times10^{8}}{1.51}=2.0\times10^{8}\ \text{m/s}$.
Check $1.51$ lies between the values for quartz and crown glass in the table — a sensible answer for ordinary glass. In a real experiment, repeat for several angles and average $\sin i/\sin r$.
⚠️ Common mistakes

1. Measuring from the surface. $i$ and $r$ are measured from the normal. A ray "at $30^\circ$ to the surface" has $i=60^\circ$.

2. Dividing the angles. Snell's law relates the sines: $\dfrac{\sin 45^\circ}{\sin 30^\circ}=1.41$, whereas $\dfrac{45}{30}=1.5$ is wrong.

3. Mixing up $n_{21}$ and $n_{12}$. $n_{21}$ is "medium 2 relative to medium 1" $=\dfrac{v_1}{v_2}=\dfrac{n_2}{n_1}$. The refractive index of water relative to glass is $\dfrac{1.33}{1.50}=0.887$, less than 1.

4. Optically denser ≠ heavier. Compare refractive indices (or speeds of light), not mass densities.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — optional: real and apparent depth

In JEE Main, NEET and JEE Advanced 2026; not in the CBSE Class 10 syllabus.

For an object at the bottom of water, viewed from almost directly above, the small-angle forms of Snell's law give a simple result: $$n=\frac{\text{real depth}}{\text{apparent depth}}.$$ A pool $2.0\ \text{m}$ deep, with $n=\tfrac43$ for water, appears only $2.0\div\tfrac43=1.5\ \text{m}$ deep. If the ray inside a denser medium meets the boundary at a large enough angle, $n_1\sin i$ would need $\sin r\gt1$: then no light leaves at all and the boundary reflects it completely (total internal reflection — the principle of optical fibres). Both ideas are developed in Class 12 Physics (Ray Optics).

Exercises

I
Ideas of refraction. Answer in one or two sentences, with a sketch where it helps.
  • 1 What is refraction, and what causes it?
  • 2 A ray of light passes from air into glass along the normal. What happens to its direction and to its speed?
  • 3 A ray passes obliquely from glass into air. Does it bend towards or away from the normal?
  • 4 Kerosene ($n=1.44$) floats on water ($n=1.33$). Which is optically denser? Which has the greater mass density?
  • 5 Why does a swimming pool look shallower than it really is?
  • 6 Why does a straight pencil, partly dipped in water at a slant, look bent at the surface?
  • 7 State the two laws of refraction.
  • 8 Why is the ray that comes out of a rectangular glass slab parallel to the ray that went in?
1 Refraction is the change in direction of light as it passes obliquely from one transparent medium into another. It happens because light travels at different speeds in different media.
2 Its direction does not change ($i=r=0^\circ$), but it slows down from about $3.0\times10^{8}$ to about $2.0\times10^{8}\ \text{m/s}$.
3 Away from the normal — it passes from a denser into a rarer medium and speeds up.
4 Kerosene is optically denser (higher $n$); water has the greater mass density (kerosene floats on it).
5 Light from the bottom bends away from the normal as it leaves the water. Traced back by the eye, the rays seem to come from a point higher than the real bottom.
6 Light from the dipped part bends away from the normal at the surface, so that part appears raised; the pencil seems to break where it enters the water.
7 (i) The incident ray, the refracted ray and the normal at the point of incidence lie in one plane. (ii) For a given colour and pair of media, $\dfrac{\sin i}{\sin r}$ is constant (Snell's law).
8 At the two parallel faces the refractions are equal and opposite: $\sin i=n\sin r$ at entry and $n\sin r=\sin e$ at exit, so $e=i$.
II
Refractive index and the speed of light. Take $c=3.0\times10^{8}\ \text{m/s}$; use the table of refractive indices where needed.
  • 1 Find the speed of light in a medium of refractive index $1.50$.
  • 2 Find the speed of light in water ($n=1.33$).
  • 3 Light travels at $1.25\times10^{8}\ \text{m/s}$ in a gemstone. Find its refractive index.
  • 4 Find the refractive index of glass ($n=1.50$) relative to water ($n=1.33$).
  • 5 Find the refractive index of water relative to glass, using the same values.
  • 6 Light travels at $2.0\times10^{8}\ \text{m/s}$ in medium A and $1.5\times10^{8}\ \text{m/s}$ in medium B. Find the refractive index of B relative to A. Which medium is optically denser?
  • 7 In which of water, kerosene and diamond does light travel fastest? In which slowest?
  • 8 Medium A has $n=1.6$ and medium B has $n=1.2$. Light goes obliquely from A into B. Does it bend towards or away from the normal? Find $n_{BA}$.
1 $v=\dfrac{3.0\times10^{8}}{1.50}=2.0\times10^{8}\ \text{m/s}$.
2 $v=\dfrac{3.0\times10^{8}}{1.33}=2.26\times10^{8}\approx2.3\times10^{8}\ \text{m/s}$.
3 $n=\dfrac{3.0\times10^{8}}{1.25\times10^{8}}=2.4$ (close to diamond).
4 $n_{gw}=\dfrac{n_g}{n_w}=\dfrac{1.50}{1.33}=1.13$.
5 $n_{wg}=\dfrac{1.33}{1.50}=0.887$ — the reciprocal of part 4.
6 $n_{BA}=\dfrac{v_A}{v_B}=\dfrac{2.0}{1.5}=1.33$; B is optically denser (light is slower there).
7 Fastest in water (smallest $n$, $1.33$); slowest in diamond ($2.42$).
8 Away from the normal (denser to rarer). $n_{BA}=\dfrac{n_B}{n_A}=\dfrac{1.2}{1.6}=0.75$.
III
Snell's law. Give angles to one decimal place and refractive indices to three significant figures. Take $n=1.00$ for air.
  • 1 Light passes from air into glass ($n=1.50$) with $i=30^\circ$. Find $r$.
  • 2 Light passes from air into water ($n=1.33$) with $i=60^\circ$. Find $r$.
  • 3 What is $r$ when $i=0^\circ$, for any pair of media?
  • 4 A ray in air with $i=45^\circ$ is refracted at $r=30^\circ$ in a liquid. Find the refractive index of the liquid.
  • 5 A ray in air with $i=50^\circ$ is refracted at $r=31^\circ$ in a plastic. Find its refractive index.
  • 6 A ray inside glass ($n=1.50$) meets a glass–air surface at $20^\circ$ to the normal. Find the angle at which it leaves.
  • 7 A ray in water ($n=1.33$) meets a glass block ($n=1.50$) lying in the water at $40^\circ$ to the normal. Find the angle of refraction in the glass.
  • 8 For a certain liquid, $i=45^\circ$ in air gives $r=30^\circ$. Find $r$ when $i=60^\circ$.
Check: enter $n_1$, $n_2$ and $i$ and compare $r$ with your answer
$r$—
$\sin r$—
bends—
1 $\sin r=\dfrac{0.500}{1.50}=0.3333$, $r=19.5^\circ$.
2 $\sin r=\dfrac{0.8660}{1.33}=0.6511$, $r=40.6^\circ$.
3 $r=0^\circ$: the ray goes straight through along the normal.
4 $n=\dfrac{\sin45^\circ}{\sin30^\circ}=\dfrac{0.7071}{0.500}=1.41$.
5 $n=\dfrac{0.7660}{0.5150}=1.49$.
6 $\sin r=1.50\times\sin20^\circ=1.50\times0.3420=0.5130$, $r=30.9^\circ$ (bent away from the normal).
7 $1.33\sin40^\circ=1.50\sin r$: $\sin r=\dfrac{1.33\times0.6428}{1.50}=0.5699$, $r=34.7^\circ$ (towards the normal: glass is denser than water).
8 $n=1.414$ (from part 4's data). $\sin r=\dfrac{0.8660}{1.414}=0.6124$, $r=37.8^\circ$.
IV
Applications and challenges.
  • 1 Laboratory. A student's glass-slab readings are $i=30^\circ,\ r=19^\circ$; $i=45^\circ,\ r=28^\circ$; $i=60^\circ,\ r=35^\circ$. Find $\dfrac{\sin i}{\sin r}$ for each and the mean refractive index of the glass.
  • 2 Timing. How long does light take to cross a glass block $6.0\ \text{cm}$ thick ($n=1.50$) along the normal? How long would it take to cross the same distance in air?
  • 3 Optical fibre. A signal travels $1000\ \text{km}$ along a glass fibre whose core has $n=1.50$. Find the travel time, ignoring the zig-zag path.
  • 4 Three slabs are equally thick: water ($n=1.33$), crown glass ($1.52$) and diamond ($2.42$). A ray strikes each at $i=45^\circ$. Which gives the largest lateral shift, and in which is the emergent ray parallel to the incident ray?
  • 5 The vanishing coin. Explain, with a ray diagram, why a coin hidden by the rim of an empty cup becomes visible when the cup is filled with water.
  • 6 Challenge. Light in air meets a flat layer of water ($n=1.33$) lying on a flat glass plate ($n=1.50$) at $i=50^\circ$. Find the angle of the ray in the water and in the glass, and show that the angle in the glass is the same as if the water were not there.
1 $\dfrac{0.500}{0.3256}=1.54$; $\dfrac{0.7071}{0.4695}=1.51$; $\dfrac{0.8660}{0.5736}=1.51$. Mean $n\approx1.52$ (crown glass).
2 In glass $v=2.0\times10^{8}\ \text{m/s}$, so $t=\dfrac{0.060}{2.0\times10^{8}}=3.0\times10^{-10}\ \text{s}$; in air $t=\dfrac{0.060}{3.0\times10^{8}}=2.0\times10^{-10}\ \text{s}$.
3 $t=\dfrac{1.0\times10^{6}\ \text{m}}{2.0\times10^{8}\ \text{m/s}}=5.0\times10^{-3}\ \text{s}=5.0\ \text{ms}$.
4 Diamond (highest $n$, so it bends the ray most inside and shifts it furthest). In all three the emergent ray is parallel to the incident ray.
5 Without water, light from the coin travelling straight to the eye is blocked by the rim. With water, light from the coin bends away from the normal at the surface; the bent ray clears the rim and enters the eye, which sees the coin (raised) along the back-traced line.
6 Water: $\sin r_w=\dfrac{0.7660}{1.33}=0.5760$, $r_w=35.2^\circ$. Glass: $1.33\sin r_w=1.50\sin r_g$ gives $\sin r_g=\dfrac{0.7660}{1.50}=0.5107$, $r_g=30.7^\circ$ — exactly what air → glass directly gives, because $n\sin\theta$ is the same in every layer.
🎯 Exam corner — board and entrance-exam practice BoardNEETJEE Main

E1. Board · 2 marks The refractive index of a glass is $1.5$. What is the meaning of this statement? Find the speed of light in this glass. ($c=3.0\times10^{8}\ \text{m/s}$)

Meaning $n=\dfrac cv=1.5$: light travels $1.5$ times faster in a vacuum than in this glass.
Working $v=\dfrac cn=\dfrac{3.0\times10^{8}}{1.5}=2.0\times10^{8}\ \text{m/s}$.
Answer $2.0\times10^{8}\ \text{m/s}$. Exam tip: refractive index has no unit — writing "$1.5\ \text{m/s}$" loses the mark.

E2. Board · 3 marks Draw a labelled diagram of the path of a ray of light through a rectangular glass slab. Mark the angles of incidence, refraction and emergence, and the lateral shift. Why is the emergent ray parallel to the incident ray?

Diagram Draw the slab, a slanting incident ray meeting the top face, and the normal there; inside, the ray bends towards the normal ($\angle r\lt\angle i$); at the bottom face draw a second normal — the ray bends away and emerges at $\angle e$. Produce the incident ray as a dashed line; the perpendicular gap between it and the emergent ray is the lateral shift. (Compare Interactive 2.)
Reason The faces are parallel, so the ray meets the second face at angle $r$. Refraction at the two faces is equal and opposite: $\sin i=n\sin r=\sin e$, hence $\angle e=\angle i$.
Answer The emergent ray is parallel to the incident ray but shifted sideways. Exam tip: draw every normal as a dashed line and put arrows on all three parts of the ray.

E3. Board · Assertion–Reason Assertion (A): A ray of light falling along the normal on a glass surface passes into the glass without bending. Reason (R): The speed of light is the same in air and in glass. (A) Both A and R are true and R explains A. (B) Both are true but R does not explain A. (C) A is true but R is false. (D) A is false but R is true.

Idea At $i=0^\circ$, Snell's law gives $\sin r=0$, so $r=0^\circ$: A is true. But light does slow down in glass ($v=\tfrac cn$), so R is false.
Answer (C). Exam tip: at normal incidence the speed still changes; only the direction does not.

E4. Board · Case study A jeweller's chart lists refractive indices: water $1.33$, kerosene $1.44$, crown glass $1.52$, diamond $2.42$. (i) Find the speed of light in diamond. [2 marks] (ii) A ray enters each material from air at the same angle of incidence. In which does it bend most? [1 mark] (iii) Find the refractive index of diamond relative to crown glass. [1 mark] ($c=3.0\times10^{8}\ \text{m/s}$)

(i) $v=\dfrac{3.0\times10^{8}}{2.42}=1.24\times10^{8}\approx1.2\times10^{8}\ \text{m/s}$.
(ii) Diamond: it has the largest $n$, so $\sin r=\dfrac{\sin i}{n}$ is smallest and the ray turns furthest towards the normal.
(iii) $n_{dg}=\dfrac{2.42}{1.52}=1.59$. Exam tip: "A relative to B" = $\dfrac{n_A}{n_B}$ — the medium named first goes on top.

E5. NEET The refractive index of glass relative to water is $\tfrac98$ and that of water relative to air is $\tfrac43$. The refractive index of glass relative to air is (A) $1.50$ (B) $1.20$ (C) $0.67$ (D) $\tfrac{27}{32}$

Idea $n_{ga}=\dfrac{n_g}{n_a}=\dfrac{n_g}{n_w}\times\dfrac{n_w}{n_a}=n_{gw}\times n_{wa}$.
Working $\tfrac98\times\tfrac43=\tfrac{36}{24}=1.50$.
Answer (A). Exam tip: relative indices multiply like a chain — the middle medium cancels.

E6. JEE Main A ray of light in air falls on a glass surface of refractive index $\sqrt3$ at an angle of incidence of $60^\circ$. Part of it is reflected and part refracted. The angle between the reflected and the refracted rays is (A) $60^\circ$ (B) $75^\circ$ (C) $90^\circ$ (D) $120^\circ$

Snell $\sin r=\dfrac{\sin60^\circ}{\sqrt3}=\dfrac{\sqrt3/2}{\sqrt3}=\tfrac12$, so $r=30^\circ$.
Geometry The reflected ray makes $60^\circ$ with the normal above the surface, i.e. $30^\circ$ with the surface; the refracted ray makes $30^\circ$ with the normal below, i.e. $60^\circ$ with the surface. Angle between them $=180^\circ-30^\circ-60^\circ=90^\circ$.
Answer (C). Exam tip: the angle between reflected and refracted rays is $180^\circ-i-r$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. Can the refractive index of a sugar solution tell you how much sugar it contains?

Why it matters. Sugar mills in India and Brazil, fruit-juice factories, honey graders and wine makers all use a small instrument called a refractometer: a drop of liquid on a prism, a glance through the eyepiece, and the sugar content is read off a scale. Hospital laboratories use the same idea to test urine and blood serum. It works because dissolving sugar in water raises its refractive index in a regular way.

What to do.

  1. Fill a thin-walled rectangular transparent box (or a flat-sided glass dish) with water. Using a laser pointer (never look into the beam) or the four-pin method, send a ray into the water at $i=40^\circ$, $50^\circ$ and $60^\circ$ and measure $r$ each time.
  2. Calculate $\dfrac{\sin i}{\sin r}$ for each and take the mean. Compare it with $1.33$.
  3. Repeat with solutions of 10, 20 and 30 grams of sugar dissolved in 100 g of solution.
  4. Plot refractive index against sugar concentration. Is the graph a straight line? Use it to find the sugar content of a packaged fruit drink.

Tools: laser pointer or pins, protractor, ruler, kitchen scale, sugar, calculator or spreadsheet.

Hand in: Your table of $i$, $r$ and $\dfrac{\sin i}{\sin r}$, your graph, your estimate for the fruit drink compared with the sugar on its label, and one sentence on the biggest source of error.

Going further. Find out what the "Brix" scale on a refractometer means and why the instrument must be used at a fixed temperature.

Instructor note: accepted values (sodium light, 20 °C) are about $1.333$ for water, $1.348$ for a 10% sucrose solution, $1.364$ for 20% and $1.381$ for 30% by mass — nearly linear, roughly $+0.0016$ per percent. With a protractor reading to $\pm1^\circ$, students typically get $n$ within about $\pm0.03$, so the 0–30% range is detectable but individual steps of 10% are only just resolved — a good discussion of measurement uncertainty.

🧪 Self-check quiz

0/5 correct

Q1. A ray of light passes obliquely from air into water. It

Q2. The refractive index of a medium is $2.0$. The speed of light in it is

Q3. Which statement about a rectangular glass slab is correct?

Q4. Medium A has $n=1.5$ and medium B has $n=1.2$. The refractive index of A relative to B is

Q5. A ray in air strikes a liquid with $i=60^\circ$ and is refracted at $r=30^\circ$. The refractive index of the liquid is

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