2022 A/L Physics Past Paper · Question 06 (English Medium)

Answer: (5) $E$

What the question asks

We want: the one ray in the figure that is drawn wrongly after the lens.

What each letter means:

$F$: principal focus of the lens.

$A$, $B$, $C$, $D$, $E$: the five rays leaving the lens.

Relevant theory

Images formed by lenses can be constructed by using two rays from the topmost point of the object, one passing parallel to the axis of the lens and the other passing through the optical centre of the lens and then locating the point where the rays meet after refraction through the lens.

The midpoint of a lens through which the rays are supposed to pass without any deviation

For all positions of the object the image is virtual, diminished, upright and forms between the focus and the lens.

Applying it to this question

The objects in the figure:

A diverging (concave) lens, with its optical centre on the axis.

An arrow on the left of the lens, with two points $F$ on the axis, one on each side.

Five rays from the tip of the arrow. After the lens they are $A$, $B$, $C$, $D$ and $E$.

Reasoning and calculation

For example, look through the small peephole in a front door. It is a diverging lens.

A visitor outside looks small, upright and much nearer than the visitor really is.

Your eye follows each ray backwards in a straight line. All the lines meet at one point, the image.

The book says this image is virtual, diminished, upright and between the focus and the lens.

So after a diverging lens, every ray drawn backwards must pass through the image of the arrow tip.

The image of the arrow tip is above the axis, between the focus $F$ and the lens.

Now check the rays one by one. Draw each one backwards.

$D$ goes through the optical centre. It is not bent, and it is a correct ray.

$C$ runs along the axis, through the optical centre. It is a correct ray.

$A$ spreads away above the axis. Drawn backwards it meets the image above the axis. It is correct.

$B$ leaves level with the axis, at the height of the image above the axis. Drawn backwards it meets the image. It is correct.

$E$ leaves the lens level, but far below the axis.

Drawn backwards it is a level line below the axis. It never reaches the image, which is above the axis.

So $E$ is not a correct ray.

Reading the options

The ray that cannot come from the image of the arrow tip is $E$.

Option (5).

Why the other options fail

Ray $A$ drawn backwards passes through the image of the arrow tip.

Ray $B$ drawn backwards passes through the image of the arrow tip.

Ray $C$ passes through the optical centre, so it is not bent.

Ray $D$ passes through the optical centre, so it is not bent.

Checks

Does it make sense?

The image of a diverging lens is upright and on the same side as the arrow.

So every ray must seem to come from a point above the axis.

A ray that stays below the axis cannot come from such a point.

Conclusion

Every correct ray, drawn backwards, passes through the image of the arrow tip. That image is above the axis.

Ray $E$ stays below the axis, so it is not correct. The answer is (5).

For a diverging lens the image is virtual, upright and small.

Draw each ray backwards. A correct ray always points back to that image.

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