2024 A/L Physics Past Paper · Question 09 (English Medium)

Answer: (3) $84\ \text{cm}$

Given information

This question is about an astronomical telescope, the tube we use to look at the Moon and the stars.

It has two lenses: the objective at the front and the eyepiece near your eye.

In normal adjustment the final image forms at infinity, very far away, so the eye looks at it relaxed.

We know the focal length of the objective, $80\ \text{cm}$, and that the telescope makes things look $20$ times bigger.

What we must find: how far apart the two lenses are.

How we find it: from the magnification we find the focal length of the eyepiece. In normal adjustment the distance between the lenses is the two focal lengths added.

The telescope is in normal adjustment. The objective focal length is $80\ \text{cm}$ and the angular magnification is $20$.

What the question asks

$f_A$ ($\text{cm}$): objective focal length.

$f_B$ ($\text{cm}$): eyepiece focal length.

$m$: angular magnification, a ratio with no unit.

$L$ ($\text{cm}$): distance between the lenses.

Relevant theory

When using telescopes it is customary for the final image to be formed at infinity. When the final image is at infinity, the telescope is said to be in "normal adjustment". The angular magnification or the magnifying power of a telescope in normal adjustment is defined as follows.

Angular magnification , (m) = α/β = tan α / tan β = (h/f_B)/(h/f_A) = |f_A/f_B|

Distance between the two lenses = |f_A| + |f_B|

What is happening: the objective brings light from the distant object to its focal plane. That image is also at the eyepiece focus. The light leaves the eyepiece parallel, so the eye views the final image at infinity.

For example, look at a distant tree through a small telescope. The front lens makes an image inside the tube. The lens near your eye lets you view that image.

In normal adjustment their focal planes meet. The distance along the tube has two parts: one focal length from each lens.

In words: add the two focal lengths.

$$L=f_A+f_B$$

Required equations

$m=\dfrac{f_A}{f_B}$

$f_B=\dfrac{f_A}{m}$

$L=f_A+f_B$

Applying it to this question

$0.05$

$\dfrac{f_B}{f_A}=\dfrac{1}{20}$

$4$

$f_B=80\times 0.05$

$84$

$L=80+4$

Reasoning and calculation

The common focus lies between the convex lenses. The lens separation is therefore greater than $80\ \text{cm}$. Only $84\ \text{cm}$ also gives the required magnification of $20$.

This gives $L=40\ \text{cm}$. Then $f_B=L-f_A=-40\ \text{cm}$. A convex eyepiece here has a positive focal length. Halving $80$ ignores the given magnification.

This subtracts the eyepiece focal length: $80-4=76\ \text{cm}$. In normal adjustment the common focal plane is between the lenses, so their positive focal lengths must add.

This would give $f_B=96-80=16\ \text{cm}$ and $m=80/16=5$. It fails the given magnification of $20$.

This treats the magnification $20$ as a length and adds it to $80\ \text{cm}$. Magnification has no unit. A separation of $100\ \text{cm}$ gives $f_B=20\ \text{cm}$ and $m=4$, not $20$.

Checks

$80/(84-80)$

From the proposed separation, $f_B=84-80=4\ \text{cm}$. Then $80/4=20$. This reverses the working and recovers the given magnification.

Both focal lengths are lengths. Adding them leaves the answer in $\text{cm}$. The magnification is dimensionless.

A magnification of $20$ needs an eyepiece with a much shorter focal length than the objective. Here $4\ \text{cm}$ is one twentieth of $80\ \text{cm}$, and both lengths are positive.

Conclusion

$L=84\ \text{cm}$

The distance between the objective lens and the eyepiece is $84\ \text{cm}$. The answer is (3).