2023 A/L Physics Past Paper · Question 20 (English Medium)

Answer: (2) Graph (2)

What the question asks

We want: the graph of the linear magnification $m$ of a real image against its image distance $V$, for a convex lens.

What each letter means:

$m$: the linear magnification. It has no unit.

$V$ ($\text{m}$): the image distance, on the horizontal axis.

$u$ ($\text{m}$): the object distance.

$f$ ($\text{m}$): the focal length of the lens. It does not change.

There are no numbers. We keep the letters, and later we check with sample values: $V = f$, $2f$ and $3f$.

Relevant theory

The lens formula, a relation connecting the object distance and the image distance with the focal length of a lens can be derived using any of the ray diagrams containing both the object and the image.

Also linear magnification (m) = height of image / Height of object = OP/IP = |v/u|

The first quote says the lens formula joins the object distance, the image distance and the focal length. The second quote says the linear magnification is $\dfrac{V}{u}$, the image distance over the object distance.

For a real image, the distances $u$ and $V$ are plain lengths on the two sides of the lens, and the lens formula reads $\dfrac{1}{u} + \dfrac{1}{V} = \dfrac{1}{f}$.

The second quote writes the image distance as a small $v$. We use the capital $V$, as the question does.

For example, think of a slide projector. Move the slide a little closer to the lens, and the picture on the far wall gets bigger. The wall has to be moved farther away to keep it sharp.

So a bigger picture goes with a bigger image distance. We now find exactly how.

Required equations

$\dfrac{1}{u} + \dfrac{1}{V} = \dfrac{1}{f}$

$m = \dfrac{V}{u}$

$\dfrac{V}{u} + 1 = \dfrac{V}{f}$

$m = \dfrac{V}{f} - 1$

Reasoning and calculation

Where does the graph exist?

A real image is formed when the object is farther than the focal length from the lens. Then the image is farther than $f$ as well, so $V$ is bigger than $f$.

So the graph begins at $V = f$.

$0$

$m = \dfrac{f}{f} - 1$

$1$

$m = \dfrac{2f}{f} - 1$

$2$

$m = \dfrac{3f}{f} - 1$

The slope and the shape:

$m = \dfrac{V}{f} - 1$ is a straight line, because $m$ goes up by the same amount for each equal step of $V$.

The slope is $\dfrac{1}{f}$, which is positive. So the line rises.

It meets the $V$ axis where $m = 0$, that is at $V = f$. It does not start at the origin.

The graph we expect:

A straight line. It starts on the $V$ axis at $V = f$, away from the origin. It rises as $V$ grows.

$v - 1$

Does it make sense?

Put a sample back: with $V = 2f$ the lens formula gives $\dfrac{1}{u} = \dfrac{1}{f} - \dfrac{1}{2f} = \dfrac{1}{2f}$, so $u = 2f$ and $m = \dfrac{V}{u} = 1$. That is the same as the line gives at $V = 2f$.

The image is the same size as the object when the object is at $2f$. That is a known fact, and it fits.

Reading the options

Now match it with the five graphs.

Graph (1) is a line from the origin. Ours starts at $V = f$, not at the origin.

Graph (2) is a straight line that starts on the $V$ axis away from the origin, and rises. It matches.

Graph (3) starts on the $m$ axis. Graph (4) is a curve, and graph (5) falls. None of these match.

Why the other options fail

Graph (1) is a line from the origin. Ours starts at $V = f$, not at the origin.

Graph (3) starts on the $m$ axis.

Graph (4) is a curve.

Graph (5) falls.

Conclusion

$m = \dfrac{V}{f} - 1$

The graph is a straight line that starts on the $V$ axis at $V = f$ and rises. That is graph (2).

The answer is (2).

For a real image, write $m = \dfrac{V}{u}$ and use $\dfrac{1}{u} + \dfrac{1}{V} = \dfrac{1}{f}$ to get $m = \dfrac{V}{f} - 1$.

That is a straight line with a positive slope, and it meets the $V$ axis at $V = f$.

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