Reflection, Refraction and Lenses: Question 7

Worked solution

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 front face of the glass panel.

Part (b): Angle of refraction at the front face

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

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

Rearranging for $\sin r$:

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

Substituting $i = 50°$ and $n = 1.50$:

$\sin r = \frac{\sin 50°}{1.50} = \frac{0.7660}{1.50} \approx 0.5107$

$r = \sin^{-1}(0.5107)$

$r \approx 30.7°$

Part (c): Angle of incidence at the back face

Inside the glass, the ray travels in a single straight line from the point where it entered the front face to the point where it strikes the back face.

Because the panel’s front and back faces are parallel to each other, the normals drawn at these two points (each perpendicular to its own face) are also parallel lines. The straight ray inside the glass crosses both of these parallel normals, acting like a transversal, so, by alternate angles, it makes the same angle with each normal.

This means the angle of incidence at the back face, measured from the normal there, equals the angle of refraction found at the front face:

$\text{angle of incidence at back face} \approx 30.7°$

Part (d): Angle of emergence at the back face

At the back face, the ray is now travelling from inside the glass out into the air, so the roles of $i$ and $r$ in the refractive-index equation swap round. Using the angle of incidence inside the glass ($30.7°$, from part (c)) to find the angle of emergence $e$ in the air:

$n = \frac{\sin e}{\sin(30.7°)} \quad\Rightarrow\quad \sin e = n \times \sin(30.7°)$

$\sin e = 1.50 \times 0.5107 \approx 0.7660$

$e = \sin^{-1}(0.7660)$

$e \approx 50°$

This is exactly the same as the original angle of incidence, $i = 50°$, at the front face.

Since the ray leaves the back face at the same angle to the normal as it entered the front face, the emergent ray travels in exactly the same direction as the original incident ray. It is only shifted sideways slightly by passing through the glass, not bent overall. Recovering the original $50°$ confirms that the restorer’s panel still has genuinely parallel faces; if the back face were tilted relative to the front face, the emergent ray would come out at some different angle instead.

Final answers

• (a) The ray bends towards the normal, because light slows down entering the optically denser glass
• (b) Angle of refraction at the front face $\approx \boxed{30.7°}$
• (c) Angle of incidence at the back face $\approx \boxed{30.7°}$, equal to (b) because the parallel faces give equal alternate angles with the ray
• (d) Angle of emergence $\approx \boxed{50°}$, equal to the original angle of incidence. Showing the emergent ray is parallel to the incident ray, confirming the panel’s faces are parallel