Reflection, Refraction and Lenses: Question 9

Syllabus 3.2.2

Multiple choice Extended 1 mark

A ray of light travelling in air strikes the flat surface of a transparent plastic block at an angle of incidence of $60°$ to the normal. The refractive index of the plastic is $1.60$ . What is the angle of refraction of the ray inside the plastic block?

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Step 1: Rearrange the refractive index equation for the angle of refraction

The refractive index relates the angle of incidence in air, $i$ , and the angle of refraction inside the material, $r$ :

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

Rearranging for $\sin r$ :

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

Step 2: Substitute the given values

With $i = 60°$ and $n = 1.60$ :

$\sin r = \frac{\sin 60°}{1.60} = \frac{0.8660}{1.60} \approx 0.5413$

Step 3: Take the inverse sine

$r = \sin^{-1}(0.5413)$

$r \approx 32.8°$

Why the other options are wrong

Option	How it arises	Error
$37.5°$	$60° \div 1.60$	Divides the angle itself by $n$ , instead of dividing $\sin i$ by $n$
$38.7°$	$\sin^{-1}(1 \div 1.60)$	Uses the critical-angle formula $\sin c = 1/n$ instead of the refraction formula
$57.2°$	$90° - 32.8°$	Measures the angle from the block’s surface instead of from the normal

Final answers

Angle of refraction $\approx \boxed{32.8°}$ , option A

Common mistakes

Dividing the angle of incidence by the refractive index directly, giving $60° \div 1.60 = 37.5°$ , instead of dividing the sines of the angles.
Confusing the refraction formula with the critical angle formula, calculating $\sin^{-1}\left(\dfrac{1}{1.60}\right) = 38.7°$ instead of $\sin^{-1}\left(\dfrac{\sin 60°}{1.60}\right)$ .
Measuring the angle from the surface of the block rather than from the normal, giving $90° - 32.8° = 57.2°$ .
Inverting the refractive index formula and attempting $\sin r = 1.60 \times \sin 60°$ , which gives a value greater than $1$ and has no valid angle.