Fig. 1.1 shows a rectangular glass block placed on a sheet of white paper. A student uses two pins on one side of the block and aligns two more pins on the ...
Question 1 Report
Fig. 1.1 shows a rectangular glass block placed on a sheet of white paper. A student uses two pins on one side of the block and aligns two more pins on the opposite side to trace a light ray through the block. The incident ray strikes face PQ at point A. The normal at A and the refracted ray inside the block are shown.
(a) Read the angle of incidence i at point A from Fig. 1.1. [1]
(b) Read the angle of refraction r inside the block at point A. [1]
(c) Calculate the refractive index n of the glass block. Show your working. [2]
(d) Using your value of n, calculate the critical angle for this glass. [2]
(e) A second ray travelling inside the glass hits face RS at an angle of 50° to the normal. State and explain what happens to this ray. [2]
(f) Suggest two reasons why the student's measured value of n may differ from the accepted value. [2]
(g) Describe how the student could use the results from several different angles to obtain a more reliable value of n. [1]
Answer Details
(a) The angle of incidence \(i\) at point A is the angle between the incoming ray and the normal at the surface PQ: [1]
\[i = 48°\](b) The angle of refraction \(r\) inside the block at point A is the angle between the refracted ray and the normal: [1]
\[r = 29°\]The ray bends towards the normal on entering the glass because glass is optically denser than air.
(c) Applying Snell's law: [2]
\[n = \frac{\sin i}{\sin r} = \frac{\sin 48°}{\sin 29°}\] \[n = \frac{0.7431}{0.4848} = 1.53\](d) The critical angle \(c\) is related to the refractive index by: [2]
\[\sin c = \frac{1}{n} = \frac{1}{1.53} = 0.654\] \[c = \sin^{-1}(0.654) = 40.8° \approx 41°\]At angles of incidence greater than 41° (when light travels from glass to air), total internal reflection occurs.
(e) The ray hits face RS at 50° to the normal. Since \(50° > 41°\) (the critical angle), total internal reflection occurs. [2]
The two conditions for TIR are both satisfied: the light is travelling from a denser medium (glass) towards a less dense medium (air), and the angle of incidence exceeds the critical angle. The ray is reflected back into the glass with no light escaping through the surface.
(f) Two reasons why the measured value may differ from the accepted value: [2]
- Parallax error when positioning the pins, leading to inaccurate marking of the ray paths.
- Difficulty in aligning the pins precisely through the block, or protractor reading errors when measuring the angles.
(g) Repeat the experiment using several different angles of incidence. Plot \(\sin i\) against \(\sin r\) and draw the best-fit straight line through the origin. The gradient of this line gives a more reliable value of \(n\) because it uses all the data points rather than relying on a single pair of angles. [1]
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