2022 A/L Physics Past Paper · Question 28 (English Medium)

Answer: (1) $8.0\ \text{cm}$

Given information

What we know:

The film is $10\ \text{cm}$ from the lens. The lens is $10\ \text{cm}$ from the container. The plant is at the far side of the water, $40\ \text{cm}$ from the container wall near the lens.

The glass walls are thin, so we ignore them. The water has refractive index $\dfrac{4}{3}$.

A clear image of the plant falls on the film. So the film is at the image position.

What the question asks

We want: the focal length $f$ of the convex lens.

What each letter means:

$n$: the refractive index of the water.

$d_{\text{real}}$ ($\text{cm}$): the real depth of the plant in the water.

$d_{\text{app}}$ ($\text{cm}$): the apparent depth of the plant, as seen from the air.

$u$ ($\text{cm}$): the object distance from the lens.

$v$ ($\text{cm}$): the image distance from the lens.

$f$ ($\text{cm}$): the focal length of the lens.

Relevant theory

OP/IP = Real depth / Apparent depth

1/v − 1/u = 1/f is the lens formula.

1/(−40) − 1/10 = 1/f

−5/40 = 1/f

f = -8 cm

For example, a coin at the bottom of a bowl of water looks closer to the top than it really is. The lens sees the plant in the same way. It sees the plant nearer than its real position.

The ratio of the real depth to the apparent depth is the refractive index of the water.

$$n = \dfrac{d_{\text{real}}}{d_{\text{app}}}$$

In words: the apparent depth is the real depth divided by the refractive index.

So the lens treats the plant as an object at the apparent depth. We add the distance between the lens and the container to find the object distance $u$.

The film holds a real image on the other side of the lens. In the book's sign convention $u$ is positive, a real image has a negative $v$, and a convex lens has a negative $f$. The size of $f$ is its focal length.

$$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$$

In words: one over the image distance minus one over the object distance is one over the focal length.

Required equations

$d_{\text{app}} = \dfrac{d_{\text{real}}}{n}$

$u = 10 + d_{\text{app}}$

$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

Applying it to this question

All distances are in $\text{cm}$, so $\dfrac{1}{u}$, $\dfrac{1}{v}$ and $\dfrac{1}{f}$ are in $\text{cm}^{-1}$. The refractive index has no unit.

$30$

$d_{\text{app}} = \dfrac{40}{4/3} = 30$

$40$

$u = 10 + 30 = 40$

$-10$

$v = -10$

$8$

$\dfrac{1}{f} = \dfrac{1}{-10} - \dfrac{1}{40} = -\dfrac{5}{40} \Rightarrow f = -8 \Rightarrow |f| = 8$

Reasoning and calculation

The minus sign only says that the lens is convex in this sign convention. Its focal length is $8.0\ \text{cm}$.

That is option (1).

Checks

$u * v / (u + v)$

Does it make sense?

The object is $40\ \text{cm}$ from the lens and the image is $10\ \text{cm}$ away. The focal length must be smaller than the image distance. $8$ is smaller than $10$, so it fits.

The plant is far from the lens compared with the film distance, so the image is small and close to the lens, a little beyond the focal length.

Conclusion

$f = 8.0\ \text{cm}$

The lens sees the plant at an apparent depth $d_{\text{app}} = 30\ \text{cm}$, so $u = 40\ \text{cm}$. The film is at $v = 10\ \text{cm}$, so $f = 8.0\ \text{cm}$.

The answer is (1).

When light leaves water into air, the object looks nearer: apparent depth is real depth divided by the refractive index.

Find the new object distance first, and only then use the lens formula.

Exam tip