IGCSE Physics 0625: Refraction, Critical Angle and Total Internal Reflection in a Fuel Sensor

An optical engineer is designing a small glass sensor that will detect the fuel level inside a car's fuel tank. Before building the sensor, she first tests how the glass bends light by shining a laser beam from air into the flat side of a rectangular block cut from the same glass.

(a) State whether the light ray bends towards or away from the normal as it enters the glass block from air, and explain this in terms of the speed of light in each material. [2]

(b) State what is meant by the critical angle for a ray of light travelling inside the glass towards a glass–air boundary. [1]

(c) State the name of the effect that occurs at a glass–air boundary when the angle of incidence inside the glass exceeds the critical angle, and state what happens to the light ray in this case. [2]

(d) The engineer now builds the sensor: a small glass prism is fixed with its tip pointing down into the fuel tank. A beam of light is shone into the prism so that it strikes the sloped inside surface of the tip at a fixed angle of incidence of $45°$ . The critical angle for the glass–air boundary is $40°$ , and the critical angle for the glass–fuel boundary is $50°$ . Use these values to explain why a detector receives a strong reflected signal when the tip is surrounded by air, but a much weaker signal when the tip is submerged in fuel. [3]

Part (a): Direction of bending on entering the glass

Light travels more slowly in glass than it does in air, because glass is optically denser than air. Whenever light slows down on crossing into an optically denser material, it bends towards the normal.

So the ray bends towards the normal as it enters the glass block.

Part (b): Definition of the critical angle

The critical angle is a special angle of incidence inside the glass, for a ray approaching a glass–air boundary. It is the angle of incidence at which the refracted ray in the air grazes exactly along the boundary. That is, the angle of incidence for which the angle of refraction is $90°$ .

Part (c): Total internal reflection

If the angle of incidence inside the glass is made greater than the critical angle, the ray can no longer refract out into the air at all. Instead, total internal reflection occurs: the entire ray is reflected back into the glass, obeying the normal law of reflection, and none of it is transmitted into the air.

Part (d): Comparing the sensor’s two boundaries

The prism tip always presents the light ray to the boundary at the same fixed angle of incidence, $45°$ . What differs is the critical angle of the boundary on the other side of the glass, air or fuel.

Tip in air: $\text{angle of incidence} = 45° > \text{critical angle} = 40°$

Since $45°$ exceeds the critical angle for the glass–air boundary, total internal reflection occurs. All of the light reflects back inside the prism towards the detector, so the detector receives a strong signal.

Tip in fuel: $\text{angle of incidence} = 45° < \text{critical angle} = 50°$

Since $45°$ is now less than the critical angle for the glass–fuel boundary, the condition for total internal reflection is not met. Most of the light instead refracts out of the prism and into the fuel, so only a weak signal (if any) returns to the detector.

This difference in signal strength is exactly what allows the sensor to detect whether the prism tip is in air or submerged in fuel.

Final answers

(a) The ray bends towards the normal, because light slows down entering the optically denser glass
(b) Critical angle $=$ the angle of incidence in the glass at which the angle of refraction in air is $\boxed{90°}$
(c) Total internal reflection. The ray is entirely reflected back into the glass, with none refracted out
(d) In air, $45° > 40°$ so total internal reflection gives a strong signal; in fuel, $45° < 50°$ so light escapes and gives a weak signal