6.1 - Refraction of Light

Refraction and Optical Density

Refraction occurs because light changes velocity when it travels between media of different optical densities. A pencil in water or seen through a glass block can appear bent.
From an optically less dense to a denser medium, light slows down and bends towards the normal: $r < i$ .
From an optically denser to a less dense medium, light speeds up and bends away from the normal: $r > i$ .
Angle of incidence, $i$ , and angle of refraction, $r$ , are measured between the respective ray and the normal.
Optical density differs from mass per unit volume. Oil floats on water because its mass density is lower, although its optical density is higher.

Refractive index, $n$ , is the ratio of light speed in vacuum to light speed in a medium:

n = \frac{c}{v}, c = 3.0 \times 1 0^{8} m s^{- 1}

Higher optical density gives a larger $n$ . The value is at least 1 and depends on light wavelength; temperature changes can also change it.
The incident ray, refracted ray and normal meet at one point and lie in the same plane.
Snell’s law relates the two media:

n_{1} sin θ_{1} = n_{2} sin θ_{2}

$n_{1}, n_{2}$ : refractive indices of the incident and refracting media; $θ_{1}, θ_{2}$ : respective angles to the normal.
For light entering a medium from air, take $n_{1} = 1$ :

n = \frac{sin i}{sin r}

At an incidence angle of $5 0^{\circ}$ from water, refraction angles in perspex, flint glass and diamond are $42. 8^{\circ}$ , $37. 9^{\circ}$ and $24. 9^{\circ}$ respectively: the higher the second medium’s index, the greater the bending towards the normal.

Inference: Refraction angle depends on incidence angle.
Hypothesis: Increasing $i$ increases $r$ .
Aim: Determine the refractive index of a glass block.
Variables: Manipulated — $i$ ; responding — $r$ ; constant — refractive index of the glass block.
Apparatus: Ray box with a single slit, protractor and ruler. Materials: Glass block, white paper and pencil. Perspex may replace glass.

Trace the block’s outline on white paper and draw a normal at incidence point $O$ .
Draw incident-ray lines at $i = 2 0^{\circ}, 3 0^{\circ}, 4 0^{\circ}, 5 0^{\circ}, 6 0^{\circ}$ .
Place the block on its outline and direct the ray along the $2 0^{\circ}$ line. Mark the emergent ray $PQ$ .
Remove the block, join $O$ to emergence point $P$ , and measure $r$ .
Replace the block and repeat for the other incidence angles. Record $i$ , $r$ , $sin i$ and $sin r$ .
Plot $r$ against $i$ and $sin i$ against $sin r$ . Use $n = sin i / sin r$ to obtain $n$ from the second graph’s gradient.

Real depth, $H$ : Distance from the actual object to the medium’s surface.
Apparent depth, $h$ : Distance from the object’s image to that surface.
Rays from a fish bend away from the normal on leaving water. Their backward extensions locate an image closer to the surface, so the fish appears shallower.
An instructor above a swimming pool sees a diver nearer the water surface; the diver sees the instructor farther away.
The relationship for a transparent medium viewed from air is:

n = \frac{H}{h}

Inference: Image position depends on object position and the refractive indices of the media.
Hypothesis: Greater real depth gives greater apparent depth.
Aim: Determine water’s refractive index by the non-parallax method.
Variables: Manipulated — $H$ ; responding — $h$ ; constant — water’s refractive index.
Apparatus: 1 000 ml beaker, ruler and retort stand with two clamps. Materials: Cork, two pins, cellophane tape and water.

Tape object pin $P_{o}$ to the base of the beaker. Arrange the ruler and adjustable comparison pin $P_{i}$ beside it using the stand and cork.
Fill with water to $H = 6.0 cm$ and view the image of $P_{o}$ from above.
Adjust $P_{i}$ vertically until it lines up with the image of $P_{o}$ at the same level.
Measure $y$ , the height of $P_{i}$ above the base. Calculate $h = H - y$ .
Repeat with $H = 7.0, 8.0, 9.0, 10.0 cm$ . Record $H$ , $y$ and $h$ .
Plot $H$ against $h$ ; its gradient gives $n = H / h$ .

r = sin^{- 1} (\frac{sin 6 0 ^{\circ}}{1.49}) = 35.5 4^{\circ}, v = \frac{3.0 \times 1 0 ^{8}}{1.49} = 2.01 \times 1 0^{8} m s^{- 1}

Apparent rise of a coin: The image is $2 H /7$ above the coin, so $h = H - 2 H /7 = 5 H /7$ and $n = H / h = 7/5 = 1.4$ .