2018 A/L Physics Past Paper · Question 37 (English Medium)

The image of "23" as a convex lens is lifted

Number 23 is written on a piece of paper placed on the flat horizontal surface of a table. A thin convex lens is held just above the number and then it is slowly taken vertically upward while looking at the image of the number through it, and keeping its optical axis vertical. Which of the following best represents the variation of the size and shape of the image as the lens is gradually moved up from the number 23?

(1) Figure (1)
(2) Figure (2)
(3) Figure (3)
(4) Figure (4)
(5) Figure (5)

Answer and explanation

Answer: (1) Figure (1)

What the question asks

We want: how the image of 23 changes as the lens goes up.

As the lens rises, the object distance $u$ grows.

Relevant theory

The object to be viewed is placed in front of the lens within its principal focus to form a virtual, magnified, upright image.

The object to be viewed is placed a little beyond the focus of the objective lens, to form a real, inverted and magnified image I₁.

Real, diminished and inverted

Required equations

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

$m = \frac{v}{u}$

Reading the options

Each option shows the images from left to right, as the lens rises.

Watch two things.

The size: large or small.

The shape: upright, turned through 180°, or swapped like in a mirror.

Reasoning and calculation

Stage 1: $u \approx 0$ , the lens on the number

$23$

The lens almost touches the paper.

It acts like a flat piece of glass.

So 23 looks the same size, and upright.

Stage 2: $0 < u < f$

$23, large$

The number is inside the focus.

The image is virtual, upright and magnified.

As $u$ comes close to $f$ , the image gets bigger.

At $u = f$ the image is very far away.

Past that, the image is real and inverted.

Inverted means turned top to bottom and left to right. That is a 180° turn.

Stage 3: $f < u < 2f$

$Turned through 180°, large$

The image is real, inverted and magnified.

So we see a large 23, turned through 180°.

Stage 4: $u > 2f$

$Turned through 180°, small$

The image is real, inverted and diminished.

The higher the lens goes, the smaller it gets.

So the order is: small upright, large upright, then large turned, small turned.

Only figure (1) shows this.

Checks

A quick check.

A real image from a lens is turned right round.

A lens never swaps only left and right. A plane mirror does that.

So figures (2) and (3), with mirror shapes, are wrong.

Conclusion

Upright and growing while $u < f$ .

Turned through 180° once $u > f$ , then shrinking.

The answer is figure (1).

For lens images, inverted means a 180° turn.

Turn the page round and you see the same shape. A mirror is different.

Exam tip

Explore

The lens is on the paper. 23 looks the same size.

The lens goes up. 23 grows. It stays upright.

At the focus the image becomes huge. It is not clear.

Beyond the focus 23 is turned right round. The higher the lens, the smaller it gets.

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