2019 A/L Physics Past Paper · Question 33 (English Medium)

Answer: (2) 9.8 cm

Given information

We know:

The slide picture is $30\ \text{mm} \times 40\ \text{mm}$.

The picture on the screen is $1.2\ \text{m} \times 1.6\ \text{m}$.

The screen is $4.0$ m from the lens.

The picture falls on a screen. So it is a real image.

What the question asks

We want:

The focal length $f$ of the projection lens.

Relevant theory

To obtain a common formula for both types of lenses and also for both types of images (real and virtual) a sign convention is applied.

All distances are measured from the optical centre of the lens. Those distances measured in the direction of incident light are considered negative while those measured against the direction of incident light are considered positive.

Use the distances as plain lengths.

$u$ is the distance of the slide from the lens. $v$ is the distance of the screen.

For a real image the lens formula is:

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

The magnification is:

$$m = \dfrac{\text{image height}}{\text{object height}} = \dfrac{v}{u}$$

With the sign convention the same law is written $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ with $v = -4.0$ m. Then $f$ comes out negative for a convex lens. Its size is the same.

Required equations

$m = \frac{\text{image height}}{\text{object height}} = \frac{v}{u}$

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

Applying it to this question

The plan:

First find the magnification from the heights.

Then find the distance $u$ of the slide.

Then put $u$ and $v$ in the lens formula.

Reasoning and calculation

$40$

$m = \frac{1.2}{0.030} = \frac{1.6}{0.040}$

$0.10$

$u = \frac{v}{m} = \frac{4.0}{40}$

$\frac{1}{f} = \frac{1}{0.10} + \frac{1}{4.0} = 10 + 0.25$

$10.25$

$\frac{1}{f} = 10.25\ \text{m}^{-1}$

$0.0976$

$f = \frac{1}{10.25}$

$9.8$

$f = 9.8\ \text{cm}$

Checks

Check the unit.

$m$ is a ratio of lengths, so it has no unit.

$\dfrac{1}{u}$ and $\dfrac{1}{v}$ are both in $\text{m}^{-1}$.

So $f$ comes out in metres. Then we change it to cm ✓.

Does it make sense?

A large magnification needs the slide just outside the focus.

Here $u = 10$ cm and $f = 9.8$ cm ✓.

Another way:

$$f = \dfrac{uv}{u + v} = \dfrac{0.1 \times 4}{4.1} = 0.0976\ \text{m}$$

$u * V / (u + V) * 100$

Conclusion

$|f| \approx 9.8\ \text{cm}$

$m = 40$, so $u = 10$ cm.

$$\dfrac{1}{f} = \dfrac{1}{0.1} + \dfrac{1}{4} \quad\Rightarrow\quad f \approx 9.8\ \text{cm}$$

So the answer is option (2).