2018 A/L Physics Past Paper · Question 38 (English Medium)

Hollow glass prism — the fluid in the cavity and the direction of the ray

The hollow glass prism with thick walls as shown in figure, is made of a material of refractive index $\mu_g$ . A ray of monochromatic light $PQ$ travelling in air is incident on the glass surface as shown in figure. In order to make the emergent ray travel along $X, Y$ and $Z$ directions respectively, the cavity of the hollow prism should be separately filled with transparent fluids of refractive index $\mu$ , where

(1) $\mu < \mu_g$ , $\mu = \mu_g$ , and $\mu > \mu_g$ respectively.
(2) $\mu > \mu_g$ , $\mu < \mu_g$ , and $\mu = 1$ respectively.
(3) $\mu = 1$ , $\mu = \mu_g$ , and $\mu < \mu_g$ respectively.
(4) $\mu = 1$ , $\mu < \mu_g$ , and $\mu > \mu_g$ respectively.
(5) $\mu = \mu_g$ , $\mu = 1$ , and $\mu = \mu_g$ respectively.

Answer and explanation

Answer: (1) $\mu < \mu_g$ , $\mu = \mu_g$ , and $\mu > \mu_g$ respectively.

What the question asks

We want: the refractive index $\mu$ of the fluid for each direction.

$X$ is up. $Y$ is straight on. $Z$ is down.

Relevant theory

When light passes from air to a medium such as glass its velocity decreases and the ray gets refracted or deviated towards the normal drawn to the interface at the point of incidence

The rule.

Into a less dense medium (smaller $\mu$ ), the ray bends away from the normal.

Into a denser medium (larger $\mu$ ), the ray bends towards the normal.

With the same $\mu$ , it does not bend.

Applying it to this question

Look at the figure.

The ray enters the glass at $Q$ . It goes straight across the glass wall.

Then it meets the left face of the cavity.

The normal to that face (the broken line) slopes down, below the ray.

The ray bends only where the glass meets the fluid.

Only the $\mu$ of the fluid changes. The glass keeps its $\mu_g$ .

Reasoning and calculation

$i$ is the angle in the glass. $r$ is the angle in the fluid.

$\mu_g \sin i = \mu \sin r$

$\sin r = \frac{\mu_g}{\mu}\sin i$

If $\mu < \mu_g$ , then $r > i$ . The ray bends away from the normal.

If $\mu = \mu_g$ , then $r = i$ . The ray goes straight on.

If $\mu > \mu_g$ , then $r < i$ . The ray bends towards the normal.

The normal is below the ray. So bending away from the normal means bending up.

$X$ (up)

Bent away from the normal. So the fluid is less dense: $\mu < \mu_g$

$Y$ (straight)

Not bent: $\mu = \mu_g$

$Z$ (down)

Bent towards the normal. So the fluid is denser: $\mu > \mu_g$

Reading the options

We get $\mu < \mu_g$ , $\mu = \mu_g$ and $\mu > \mu_g$ .

That is option (1).

Why the other options fail

This option says: $\mu > \mu_g$ for $X$ .

But a denser fluid bends the ray towards the normal. Then it goes down to $Z$ , not up to $X$ .

So it is wrong.

This option says: $\mu < \mu_g$ for $Z$ .

But a less dense fluid bends the ray up, to $X$ , not down to $Z$ .

This option says: $\mu < \mu_g$ for $Y$ .

But the ray goes straight on only when $\mu = \mu_g$ .

This option says: $\mu = \mu_g$ for $X$ and for $Z$ .

But with $\mu = \mu_g$ the ray does not bend at all. It goes to $Y$ .

Checks

Does it make sense?

Air has $\mu = 1$ , less than $\mu_g$ .

So an empty cavity also sends the ray up to $X$ . That fits $\mu < \mu_g$ .

Conclusion

$X$ : $\mu < \mu_g$ .

$Y$ : $\mu = \mu_g$ .

$Z$ : $\mu > \mu_g$ .

The answer is option (1).

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