2024 A/L Physics Past Paper · Question 25 (English Medium)

Answer: (2) concave, $50\ \text{cm}$

Given information

This question is about two thin lenses placed together, touching each other.

Together they act like one lens of power $+3\ \text{D}$. One of them is a convex lens we know.

What we must find: what kind of lens the other one is, and its focal length.

How we find it: powers of lenses in contact simply add. Take the known power away from the total, then turn the power back into a focal length.

Combined power $P=+3\ \text{D}$. The known lens is convex with $f_1=20\ \text{cm}=0.20\ \text{m}$.

What the question asks

$P$ ($\text{D}$): power of the two lenses together.

$P_1$, $P_2$ ($\text{D}$): power of the convex lens and of the other lens.

$f_1$, $f_2$ ($\text{m}$): their focal lengths.

Relevant theory

∴ Power of a lens P = 1/f

Sign convention :The power of a convex lens is positive while that of a concave lens is negative. Hence according to the Cartesian sign convention the power can be expressed as ,

From 9.8 and 9.9, 1/f = 1/f₁ + 1/f₂

What is happening: the convex lens alone bends light strongly, with $+5\ \text{D}$.

The pair bends light less, only $+3\ \text{D}$. So the second lens must undo part of the bending.

A lens that undoes bending spreads light out. It has a negative power, so it is concave.

For example, think of an optician testing your eyes.

They slide a strong reading lens into the frame. Everything looks too close and blurry.

Then they drop a weak spreading lens in front of it. The two together are just right.

In words: putting lenses together adds their powers, and a spreading lens takes some power away.

Required equations

$P=\dfrac{1}{f}$

$P=P_1+P_2$

$P_2=P-P_1$

Applying it to this question

$+5$

$P_1=\dfrac{1}{0.20}$

$-2$

$P_2=3-5$

$f_2=\dfrac{1}{P_2}=\dfrac{1}{-2}$

$-0.5$

$f_2$

Reasoning and calculation

$P_2$ is negative, so the other lens is concave. Its focal length is $0.5\ \text{m}=50\ \text{cm}$. This is option (2).

A convex lens of $50\ \text{cm}$ has $+2\ \text{D}$. With the $+5\ \text{D}$ lens the pair would have $+7\ \text{D}$, not $+3\ \text{D}$.

A convex lens of $12.5\ \text{cm}$ has $+8\ \text{D}$. This adds $3$ and $5$ instead of taking $5$ away from $3$. The pair would have $+13\ \text{D}$.

A concave lens of $12.5\ \text{cm}$ has $-8\ \text{D}$. With the $+5\ \text{D}$ lens the pair would have $-3\ \text{D}$. The sign of the answer is the wrong way round.

A concave lens of $10\ \text{cm}$ has $-10\ \text{D}$. With the $+5\ \text{D}$ lens the pair would have $-5\ \text{D}$, so it would spread light out, not bring it together.

Checks

$1/(P-1/f1)$

$P_1+P_2=(+5)+(-2)$

$P=+3\ \text{D}$

$1/P$ is $1$ divided by $\text{D}=\text{m}^{-1}$, so $f_2$ comes out in $\text{m}$, a length.

The concave lens is weaker than the convex one ($2\ \text{D}$ against $5\ \text{D}$), so the pair still brings light together, as $+3\ \text{D}$ says.

Conclusion

$P_2=-2\ \text{D},\ f_2=50\ \text{cm}$

The other lens is concave with focal length $50\ \text{cm}$. The answer is (2).