haya.Learning made clear
Unit IX · Ray optics and optical instruments
10 min readConcept sheet1-mark MCQ, 2-mark questions on Snell's law and refractive index, and a 3-mark question on apparent depth

Refraction of light: Snell's law and apparent depth

Light bends at a boundary because its speed changes. Snell's law: sin i / sin r = n₂₁. So water looks shallower: apparent depth = real depth ÷ n.

Snell's lawsin i / sin r = n₂₁, where i = angle of incidence, r = angle of refraction and n₂₁ = refractive index of medium 2 with respect to medium 1
Refractive indexn = c/v (c = speed of light in vacuum, v = speed in the medium). It has no unit. n₂₁ = n₂/n₁ = v₁/v₂
Useful relationsn₁₂ = 1/n₂₁ and n₃₂ = n₃₁ × n₁₂
Apparent depth (seen along the normal)h′ = h/n, where h = real depth and n = refractive index of the liquid or slab

Laws of refraction

Refraction is the bending of light when it passes at an angle from one medium into another. It happens because light travels at different speeds in different media. (1) The incident ray, the refracted ray and the normal at the point of incidence lie in one plane. (2) For a given pair of media, sin i / sin r is a constant, n₂₁. This is Snell's law. If n₂₁ > 1, medium 2 is optically denser: light slows down and bends towards the normal (r < i). If n₂₁ < 1, light speeds up and bends away from the normal. Optical density is not mass density: turpentine is optically denser than water, yet it floats on water. A ray along the normal (i = 0) goes straight through without bending.

Results you will use

  • Refractive index from speeds: n₂₁ = v₁/v₂. The absolute refractive index is n = c/v: about 1.5 for glass and 1.33 for water.
  • Reversibility: n₁₂ = 1/n₂₁. For light going from glass into air, n(air with respect to glass) = 1/1.5 = 2/3.
  • Chain rule: n₃₂ = n₃₁ × n₁₂, for three media 1, 2 and 3.
  • Glass slab: the emergent ray is parallel to the incident ray but shifted sideways. This lateral shift grows with the thickness of the slab and with the angle of incidence.
  • What changes: the speed and the wavelength change on refraction. The frequency does not.

Derivation: real depth and apparent depth

An object O lies at depth h below the surface of water of refractive index n. You look at it from directly above, so the rays stay close to the normal.

  1. 1
    A ray from O meets the surface at A, a small distance x from the normal through O. In the water it makes an angle r with the normal.
  2. 2
    In air it bends away from the normal, at angle i. Traced backwards, it meets the normal at I, at depth h′. I is the image you see.
  3. 3
    From the figure: tan r = x/h and tan i = x/h′.
  4. 4
    Snell's law, from water into air: n sin r = 1 × sin i.
  5. 5
    For near-normal viewing the angles are small, so sin ≈ tan: n x/h = x/h′.
  6. 6★
    So, h′ = h/n, that is, real depth / apparent depth = n. The object seems raised by h − h′ = h(1 − 1/n).

★ key step

How to draw the apparent depth diagram

  • Draw a horizontal line for the water surface. Shade the water below it, and write 'water, n' below and 'air' above.
  • Draw a vertical dotted normal. Mark the object O on it, a depth h below the surface.
  • Draw a ray from O to a point A on the surface, close to the normal. Draw the normal at A and mark the angle r.
  • In air, draw the refracted ray bending away from the normal at A. Mark the larger angle i, and draw an eye where the ray goes.
  • Extend the refracted ray backwards as a dotted line until it meets the vertical normal at I, above O.
  • Label the real depth h (surface to O) and the apparent depth h′ (surface to I). Put arrows on the real rays.

Worked example: Snell's law and apparent depth

Problem

(a) Light in air strikes a glass surface at 60°. The glass has n = √3. Find the angle of refraction and the speed of light in the glass (c = 3 × 10⁸ m s⁻¹). (b) A coin lies at the bottom of a bucket filled with water to a depth of 20 cm (n = 4/3). How deep does it appear from directly above, and by how much is it raised? (c) A mark under a glass slab 6.0 cm thick (n = 1.5) is viewed from above. Find its apparent depth.

Working
  1. (a) sin r = sin i / n = sin 60° ÷ √3 = (√3/2) ÷ √3 = 1/2, so r = 30°.
  2. v = c/n = (3 × 10⁸ m s⁻¹) ÷ √3 = √3 × 10⁸ m s⁻¹ ≈ 1.73 × 10⁸ m s⁻¹.
  3. (b) Apparent depth h′ = h/n = 20 cm ÷ (4/3) = 20 × 3/4 = 15 cm.
  4. The coin is raised by h − h′ = 20 cm − 15 cm = 5 cm.
  5. (c) h′ = 6.0 cm ÷ 1.5 = 4.0 cm, so the mark seems raised by 2.0 cm.
Answer

(a) r = 30°, v ≈ 1.73 × 10⁸ m s⁻¹. (b) 15 cm deep, raised by 5 cm. (c) 4.0 cm

Activity: refractive index of a glass slab and of a liquid

Two practical experiments that use the idea of real and apparent depth.

  1. 1
    Glass slab (travelling microscope): Make a small ink cross on white paper. Focus the microscope on it and note the reading R₁.
  2. 2
    Place the glass slab over the cross. The cross now looks raised. Raise the microscope to focus on it again, and note R₂.
  3. 3
    Sprinkle a little chalk dust (or lycopodium powder) on top of the slab. Focus on the dust and note R₃.
  4. 4
    Real thickness = R₃ − R₁ and apparent thickness = R₃ − R₂. So n = (R₃ − R₁)/(R₃ − R₂). Result: about 1.5 for ordinary glass.
  5. 5
    Liquid (concave mirror): Lay the concave mirror flat, facing up. Move a pin up or down above it until the pin and its inverted image coincide with no parallax. The pin is then at C, a height R above the pole.
  6. 6
    Pour a thin layer of the liquid into the mirror and repeat. The pin now coincides with its image at a smaller height R′.
  7. 7
    For a thin layer of liquid, n = R/R′. Result: for water, n comes out close to 1.33.
In real life

A temple tank or a village pond looks shallower than it really is, so never judge its depth by eye before stepping in. A coin in a glass of water looks raised for the same reason.

Exam angle

1 marks: “State Snell's law of refraction.” For a given pair of media, the ratio sin i / sin r is a constant, n₂₁; the incident ray, the refracted ray and the normal lie in one plane.

2 marks: “Light passes from air into glass (n = 1.5) at an angle of incidence of 30°. Find sin r and the speed of light in the glass (c = 3 × 10⁸ m s⁻¹).” sin r = sin 30°/1.5 = 0.5/1.5 = 1/3, so r ≈ 19.5°. v = c/n = 3 × 10⁸ ÷ 1.5 = 2 × 10⁸ m s⁻¹.

3 marks: “Explain with a ray diagram why a coin at the bottom of a tank of water appears raised. Show that real depth / apparent depth = n for near-normal viewing.” Rays from the coin bend away from the normal at the surface; traced back, they meet at I above O; tan r = x/h and tan i = x/h′; n sin r = sin i; for small angles, h/h′ = n.

Remember it

Fast to slow, bend to the normal. Light entering a denser medium slows down and bends towards the normal.