Refraction of light

Dip a pencil at an angle into a glass of water and it appears to snap sharply at the surface. The pencil has not broken — the light carrying its image to your eye has changed direction on crossing from water into air. This bending of light as it passes from one transparent medium into another is called refraction, and unlike the sudden "trick" it looks like, it obeys two precise, testable laws. This topic opens the second half of the CBSE Class 10 chapter Light — Reflection and Refraction, shifting attention from mirrors to the boundary between two transparent media. You will learn why refraction happens, how to predict which way a ray bends, and how to use Snell's law to calculate angles — the foundation for glass slabs, refractive index, and lenses later in the chapter.

The core insight

Light bends at the boundary between two transparent media because its speed changes there, and the sines of the angle of incidence and the angle of refraction stay in a fixed ratio for that pair of media, called Snell's law.

The core idea

Why refraction happens. Light travels at different speeds in different media — fastest in vacuum, a little slower in air, slower still in water, and slower yet in glass. When a ray meets the boundary between two such media at an angle, one edge of the wavefront changes speed before the other, which swings the direction of travel. This is exactly analogous to a car's wheels turning when one side rolls onto softer ground while the other is still on the road.

Which way does it bend. Two simple rules cover every case:

  • Light bends towards the normal when it goes from an optically rarer medium (like air) into an optically denser medium (like water or glass) — it slows down, and the angle from the normal decreases.
  • Light bends away from the normal when it goes from a denser medium into a rarer one — it speeds up, and the angle from the normal increases.

A ray that strikes the boundary exactly along the normal (i=0°i = 0°) passes straight through undeviated; bending needs an angled approach.

The two laws of refraction. First, the incident ray, the refracted ray, and the normal at the point of incidence all lie in the same plane. Second, for a given pair of media and a given colour of light, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant — Snell's law:

sin⁡isin⁡r=n21\frac{\sin i}{\sin r} = n_{21}

where: ii is the angle of incidence, measured in the first medium; rr is the angle of refraction, measured in the second medium; and n21n_{21} is the refractive index of the second medium with respect to the first, a constant fixed by the pair of media (and, strictly, the colour of light). This same constant also equals the ratio of the speeds of light in the two media:

n21=v1v2n_{21} = \frac{v_1}{v_2}

where: v1v_1 is the speed of light in the first medium and v2v_2 is the speed of light in the second medium — a denser medium has a smaller vv, and so a larger n21n_{21} for light entering it. Note carefully that, unlike the law of reflection, i≠ri \neq r here except at normal incidence; the two angles are tied together only through the constant ratio of their sines, never by direct equality.

Figure
A ray bending towards the normal on entering a denser medium from a rarer one, with the angle of incidence larger than the angle of refraction
Refraction: rarer to denser bends the ray towards the normal air (rarer) glass (denser) normal i = 45° r ≈ 28°
Derivation ASolving Snell's law for the angle of refraction

The goal is to turn the ratio form of Snell's law, together with a known angle of incidence and a known refractive index for the pair of media, into a numerical angle of refraction — a rearrangement every student must be able to do fluently and quickly, using only the definition of sine and its inverse.

Step 1. Write Snell's law with the known angle of incidence ii and refractive index n21n_{21} as the only inputs: sin⁡isin⁡r=n21\dfrac{\sin i}{\sin r} = n_{21}.

Step 2. Isolate sin⁡r\sin r by cross-multiplying and dividing: sin⁡r=sin⁡in21\sin r = \dfrac{\sin i}{n_{21}}. This single step is the entire algebra — the common error is to divide ii itself by n21n_{21} instead of dividing sin⁡i\sin i, which silently gives a wrong angle.

Step 3. Recover rr by taking the inverse sine of both sides: r=sin⁡−1 ⁣(sin⁡in21)r = \sin^{-1}\!\left(\dfrac{\sin i}{n_{21}}\right). Because n21>1n_{21} > 1 for a rarer-to-denser crossing, sin⁡r<sin⁡i\sin r < \sin i, so r<ir < i — the ray bends towards the normal, matching the qualitative rule above without any extra reasoning.

Step 4. Check the reverse crossing. Going from the denser medium back into the rarer one uses the reciprocal index, n12=1/n21n_{12} = 1/n_{21}, so sin⁡i′=n21sin⁡r′\sin i' = n_{21} \sin r' and the new angle is larger than the one it started from — the ray bends away from the normal, consistent with light speeding up.

Worked example

A ray of light travelling in air strikes the flat surface of a glass slab at an angle of incidence of 45°45°. Given sin⁡45°=0.71\sin 45° = 0.71 and the refractive index of glass with respect to air n21=1.5n_{21} = 1.5, find the angle of refraction and state the direction of bending.

Step 1. Apply Snell's law: sin⁡isin⁡r=n21=1.5\dfrac{\sin i}{\sin r} = n_{21} = 1.5.

Step 2. Rearrange for sin⁡r\sin r: sin⁡r=sin⁡i1.5=0.711.5≈0.473\sin r = \dfrac{\sin i}{1.5} = \dfrac{0.71}{1.5} \approx 0.473.

Step 3. Convert to an angle: since sin⁡28°≈0.469\sin 28° \approx 0.469 is close to 0.4730.473, r≈28°r \approx 28°.

Step 4. Compare r≈28°r \approx 28° with i=45°i = 45°: because r<ir < i, the ray bends towards the normal, as expected for light passing from air (rarer) into glass (denser). Answer: r≈28°r \approx 28°, bending towards the normal.

Competitive stretch

Using n21=vair/vglass=1.5n_{21} = v_{\text{air}}/v_{\text{glass}} = 1.5 and the speed of light in air vair≈3.0×108 m/sv_{\text{air}} \approx 3.0 \times 10^{8}\ \text{m/s}, the speed of light inside this glass is vglass=vair/n21=(3.0×108)/1.5=2.0×108 m/sv_{\text{glass}} = v_{\text{air}}/n_{21} = (3.0\times10^{8})/1.5 = 2.0\times10^{8}\ \text{m/s} — a third slower than in air, which is why the ray bent towards the normal in the first place.

Common traps

  • Confusing the direction of bending. Identify the denser medium first: rarer-to-denser bends towards the normal (angle decreases); denser-to-rarer bends away (angle increases).
  • Applying Snell's law to angles instead of their sines. The constant ratio is sin⁡i/sin⁡r\sin i / \sin r, not i/ri/r — a very common arithmetic slip.
  • Thinking light bends throughout a medium. Bending happens only at the boundary; light travels in a straight line within any single uniform medium.
  • Forgetting normal incidence. A ray along the normal (i=0°i = 0°) passes straight through with r=0°r = 0° too — no bending, even though the speed still changes.

Ask the tutor: If the towards-versus-away-from-the-normal rule still feels shaky, ask the AI tutor to run a few more mixed air-water-glass examples with you until it clicks.

Why does light bend at all instead of just slowing down in a straight line?

Because the ray usually meets the boundary at an angle, not head-on. Picture a wavefront as a wide marching line: the part of it that reaches the denser medium first slows down while the rest is still moving at the old speed, so the line pivots, like a row of soldiers wheeling when one end steps onto muddy ground. Only a ray travelling exactly along the normal has its whole wavefront change speed at once, so it passes straight through with no bending.

How do I remember whether the ray bends towards or away from the normal?

Link it to speed. Entering a denser medium always slows light down, and slowing down pulls the ray closer to the normal — think of it as light "hugging" the boundary less. Entering a rarer medium speeds light up and pushes the ray farther from the normal. If you know which medium is denser (usually stated, or inferable from the refractive index being greater than 1), the bending direction follows immediately without needing to calculate anything.

Is the refractive index the same for every colour of light?

Not exactly, though CBSE Class 10 problems usually treat it as one fixed number for a given pair of media. In reality, refractive index depends slightly on the wavelength (colour) of light, which is why white light splits into colours on refraction through a prism — a topic that comes later. For the angle-of-refraction calculations in this section, you can safely use the single quoted value of n21n_{21} unless the question tells you otherwise.

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