2023 A/L Physics Past Paper · Question 42 (English Medium)

Answer: (4) $\dfrac{4}{7}h$

What the question asks

We want: the height $y$ of the water, when the container looks half full.

What each letter means:

$h$ ($\text{m}$): the height of the container.

$y$ ($\text{m}$): the height of the water in it.

$n$: the refractive index of water, $\dfrac{4}{3}$.

$d$ ($\text{m}$): the apparent depth of the water, seen from above.

Relevant theory

∴ ₐn_b = OP/IP = Real depth / Apparent depth

The quote says that, for an eye looking straight down, the refractive index is the real depth divided by the apparent depth. So water of refractive index $n$ and real depth $y$ looks $n$ times shallower: its apparent depth is $\dfrac{y}{n}$.

For example, a swimming pool 2 m deep looks about 1.5 m deep when you look straight down into it.

The bottom looks nearer, but the top of the water stays where it is.

Required equations

$d = \dfrac{y}{n} = \dfrac{3y}{4}$

$h - y$

$\dfrac{3y}{4} = h - y$

$y = \dfrac{4}{7}h$

Reasoning and calculation

What the eye sees from above:

The top of the water is at the height $y$ from the bottom, so the air above it is $h - y$ deep. The eye sees this air where it really is.

The water below is seen shallower, by the refractive index. Its apparent depth is $\dfrac{y}{n} = \dfrac{3y}{4}$.

$\dfrac{3y}{4}$

$d = \dfrac{y}{n} = y \times \dfrac{3}{4}$

$h - y$

The container looks half full:

The container looks $h - y + \dfrac{3y}{4}$ deep in total, with the air on top and the water below. Half full means the water takes half of that, so the water looks as deep as the air.

This gives $\dfrac{3y}{4} = h - y$. Collect the $y$ terms: $\dfrac{7y}{4} = h$.

$\dfrac{7y}{4} = h$

$\dfrac{3y}{4} = h - y$

$\dfrac{4}{7}h$

$y = \dfrac{4h}{7}$

$4$

$y = \dfrac{4}{7} \times 7$

$3$

$\dfrac{3}{4} \times 4$

$h - y = 7 - 4$

$(h - y) / (3 * y / 4)$

Reading the options

Now compare $\dfrac{4}{7}h$ with the five options.

Option (4) is $\dfrac{4}{7}h$. It matches.

Option (3), $\dfrac{h}{2}$, is the answer if the light were not bent. The others, $\dfrac{h}{4}$, $\dfrac{h}{3}$ and $\dfrac{3h}{4}$, do not make the air and the apparent water equal.

Checks

Check the units. Both $y$ and $h$ are lengths, and $\dfrac{4}{7}$ is a pure number. So $y = \dfrac{4}{7}h$ is a length, as it must be.

The refractive index $n$ is a pure number too, so $\dfrac{y}{n}$ is also a length.

Does it make sense?

If $n = 1$, the water would not bend the light. Then $y = h - y$ and $y = \dfrac{h}{2}$, the plain half. The formula $\dfrac{y}{n} = h - y$ gives this.

With $n = \dfrac{4}{3}$ the water looks shallower, so more water, $y = \dfrac{4}{7}h$ which is a little more than half, is needed to look half full.

Conclusion

$y = \dfrac{4}{7}h$

The container looks half full when the height of the water is $y = \dfrac{4}{7}h$.

The answer is (4).

The quote gives the real depth and the apparent depth. The step that the container looks half full when the water looks as deep as the air above it is our reading of the question.

Teacher's note

Look from above: the air stays where it is, and the water looks shallower by the refractive index. Make the two depths equal for a half full look.

Exam tip