2022 A/L Physics Past Paper · Question 33 (English Medium)

Answer: (3) $n < \dfrac{3\sqrt{3}}{4}$

Given information

What we know:

The prism has the angle $60^\circ$ at $A$ and a right angle at $C$.

The ray hits the face $AC$ at right angles.

The glass has refractive index $\dfrac{3}{2}$. The liquid layer on $AB$ has refractive index $n$.

The question has no number for $n$. So we keep the letter and check the result with $n = 1.2$ and $n = 1.4$.

What the question asks

We want: the values of $n$ for which the ray is totally internally reflected at the face $AB$.

What each letter means:

$n$: the refractive index of the liquid layer.

$i$ ($^\circ$): the angle of incidence of the ray on the face $AB$, measured from the normal.

$c$ ($^\circ$): the critical angle for the glass and the liquid.

Relevant theory

What is happening:

The ray goes into the glass through $AC$ along the normal, so it does not bend. It runs straight on to the face $AB$.

At $AB$ the glass meets the liquid, and the ray arrives at a slant.

When $n$ is small, the liquid is much less dense optically than the glass. The ray cannot get out at that slant, and all of it bounces back into the glass, as if $AB$ were a mirror.

When $n$ is made bigger, closer to the glass, a ray starts to escape into the liquid and the reflected ray gets weaker.

So there is a largest $n$ that still keeps all the light inside. We have to find it.

(2) For all rays of light of a given colour passing through two given media, the sine of the angle of incidence bears a constant ratio to the sine of the angle of refraction.

(4) When the angle of incidence exceeds the critical angle, the interface acting as a mirror reflects the ray completely back into the dense medium. This phenomenon is called 'total internal reflection'.

Light should pass from a dense medium to a less dense (or rare) medium.

The angle of incidence should exceed the critical angle.

First, the angle of incidence on $AB$. The angles of a triangle add up to $180^\circ$. So the angle at $B$ is $180^\circ - 60^\circ - 90^\circ = 30^\circ$.

The ray enters $AC$ at right angles to it. The side $CB$ is also at right angles to $AC$. So the ray runs parallel to $CB$, and it makes the same angle with $AB$ as $CB$ does, which is $30^\circ$.

The normal to $AB$ is at $90^\circ$ to the face. So the angle of incidence is:

$$i = 90^\circ - 30^\circ = 60^\circ$$

In words: the angle of incidence is the angle from the normal, not the angle from the face.

Next, the limit. At the critical angle the refracted ray just runs along the face, at $90^\circ$ to the normal. For glass to liquid the ratio of the sines is $\dfrac{n}{3/2}$.

$$\sin c = \dfrac{2n}{3}$$

In words: the sine of the critical angle equals $n$ divided by the refractive index of the glass.

Total internal reflection needs the angle of incidence to be more than $c$. So $i > c$, and then $\sin 60^\circ > \sin c$.

For example, a swimmer under water looks up at the surface at a slant, far from straight up. The underside of the surface shines like a mirror, because the light cannot get out of the water at that slant.

The ray in the prism is like that light. It meets $AB$ at a big slant, so it can be held inside.

Required equations

$B = 180^\circ - 60^\circ - 90^\circ = 30^\circ$

$i = 90^\circ - 30^\circ = 60^\circ$

$\sin c = \dfrac{2n}{3}$

$\sin 60^\circ > \dfrac{2n}{3}$

$n < \dfrac{3}{2}\sin 60^\circ = \dfrac{3\sqrt{3}}{4}$

Applying it to this question

A sine has no unit. The refractive indices have no unit. So $n$ has no unit and the limit $\dfrac{3\sqrt{3}}{4}$ is a plain number.

$30$

$B = 180^\circ - 60^\circ - 90^\circ$

$60$

$i = 90^\circ - 30^\circ$

$1.299$

$n < \dfrac{3}{2} \times \sin 60^\circ = 1.5 \times 0.866$

Reasoning and calculation

The ray is totally reflected when $n < \dfrac{3\sqrt{3}}{4} \approx 1.30$.

That is option (3).

Checks

$n * s$

Does it make sense?

Try $n = 1.2$. Then $\sin c = \dfrac{2 \times 1.2}{3} = 0.8$, so $c \approx 53^\circ$. The ray comes in at $60^\circ$, which is more than $53^\circ$. It is totally reflected.

Try $n = 1.4$. Then $\sin c = \dfrac{2 \times 1.4}{3} = 0.93$, so $c \approx 69^\circ$. The ray comes in at $60^\circ$, which is less than $69^\circ$. It passes into the liquid.

So a smaller $n$ gives total internal reflection, as option (3) says.

Conclusion

$n < \dfrac{3\sqrt{3}}{4}$

The angle of incidence on $AB$ is $60^\circ$, so total internal reflection needs $n < \dfrac{3\sqrt{3}}{4}$.

The answer is (3).

The angle of incidence is measured from the normal to the face. Work it out from the angles of the triangle before you use any formula.

Total internal reflection needs light going from the denser to the less dense medium, so the answer is a limit below which $n$ must stay.

Exam tip