2023 A/L Physics Past Paper · Question 15 (English Medium)

Answer: (3) $15\ \text{cm}$

Given information

What we know:

The convex lens has focal length $f_A = 20\ \text{cm}$.

The concave lens has focal length $f_B = 5\ \text{cm}$.

A parallel beam goes into the convex lens, and a parallel beam comes out of the concave lens.

What the question asks

We want: the distance $d$ between the two lenses.

What each letter means:

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

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

$v_1$ ($\text{cm}$): the distance of the first image from the convex lens.

$x$ ($\text{cm}$): the distance of that first image behind the concave lens.

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

Relevant theory

The objective forms a real, inverted and magnified image of a distant object at its secondary principal focus.

The distance from the optical centre to each principal focus is called the focal length of the lens. For any lens the two focal lengths are equal to each other.

Similarly, principal foci exist for concave lenses as follows.

However, due to the lens B, the rays from A get refracted through B to form the final image I for which the first image I₁ can be considered as the object.

This would take place when the principal focus of the eye-piece coincides with the first image I₁ formed by the objective.

The first quote says that a lens brings the beam from a very far object together at its focus. A parallel beam is like light from a very far object.

The fourth quote says that the image from the first lens becomes the object for the second lens.

The last quote shows when the beam comes out parallel. That happens when the image from the first lens sits at the focus of the second lens.

A concave lens spreads a beam out, and its rays seem to come from its focus. Light can also run the other way. Rays that are heading towards that focus come out of the concave lens as a parallel beam. That focus lies behind the concave lens.

For example, think of a guide at a stadium gate. The first guide herds the crowd towards a lamp-post ahead. A second guide stands just before the lamp-post and turns the crowd into straight lines.

The second guide must stand at the right distance from the lamp-post. Then the crowd comes out in straight lines.

Required equations

$v_1 = f_A$

$x = f_B$

$v_1 = d + x$

$d = v_1 - x$

Applying it to this question

All the lengths are in $\text{cm}$, so we can use them as they are.

$20$

$v_1 = f_A$

$5$

$x = f_B$

$15$

$d = v_1 - x = 20 - 5$

Reasoning and calculation

The first image is $20\ \text{cm}$ from the convex lens.

It must be $5\ \text{cm}$ behind the concave lens. So the concave lens must stand $20 - 5 = 15\ \text{cm}$ from the convex lens. That is option (3).

Checks

$fa - fb$

The unit: $\text{cm}$ minus $\text{cm}$ is $\text{cm}$. A distance, as it should be.

Does it make sense?

Put the answer back: with $d = 15\ \text{cm}$ the first image is $20 - 15 = 5\ \text{cm}$ behind the concave lens. That is its focal length, so the beam leaves parallel.

The distance $d$ is less than $20\ \text{cm}$, so the concave lens stands before the point where the beam would meet. That is what we need.

Conclusion

$d = 15\ \text{cm}$

The distance is $d = f_A - f_B = 20 - 5 = 15\ \text{cm}$.

The answer is (3).

Follow the beam. The first lens makes an image at its focus. For a parallel beam out, that image must sit at the focus of the concave lens, behind it.

Then $d$ is the first focal length minus the second. Check by putting it back: the first image should be exactly one focal length of the concave lens behind it.

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