Class 12 Lens Formula Practice: Signs, Images and Magnification
Use one sign convention from the first line to the last. Here light travels left to right and distances to the right of the lens are positive. A real object on the left therefore has a negative object distance, even though its distance from the lens is a positive length.
Curriculum checked 2 October 2026 · original practice for selected skills.
Before you start
You will need: Reciprocal fractions, signed distances and principal rays.
Attempt each task before opening a hint. If you need help, reveal the first hint, then the setup, then compare your written steps with the solution. Finish with the retests without referring back.
Three ideas to keep in view
For a thin lens in air using this Cartesian convention, 1/f=1/v−1/u. Converging lenses have positive focal length.
Magnification is v/u. Its sign describes orientation relative to the object; its magnitude compares image height with object height.
A ray parallel to the axis passes through the far focus after a converging lens. A ray through its optical centre continues undeviated in the thin-lens approximation.
Worked-example ray sketch: an object 45 cm left of a converging lens with f=15 cm forms a real, inverted image 22.5 cm to the right, half the object's height. Two principal rays locate the same image point.
Try the worked example first
P12-LENS-W1Assign signsFocused practice
An object is 45 cm left of a converging thin lens of focal length 15 cm. Light travels left to right. Find v and magnification.
Show your working
Need a starting hint?
Set u=−45 cm and f=+15 cm.
Still stuck? Reveal the setup
Rearrange to 1/v=1/f+1/u.
Show worked solution+
1
Find image distance
Add signed reciprocals.
v1=151−451=452
2
Interpret magnification
v=22.5 cm and v/u=−0.5: real image to the right, inverted and half-height.
Answer: v=+22.5 cm; m=−0.5
Before moving on: can you explain why your method works, as well as give the answer?
Your turn: three different checks
Write a method as well as an answer. The tasks change the reasoning, not just the numbers.
P12-LENS-Q1Inside the focusFocused practice
An object is 10 cm left of a converging lens with f=+20 cm. Find v and describe the image.
Show your working
Need a starting hint?
The object is inside the focal distance.
Still stuck? Reveal the setup
1/v=1/20−1/10.
Show worked solution+
1
Calculate
1/v=−1/20, so v=−20 cm.
2
Interpret
m=(−20)/(−10)=+2: virtual, upright and enlarged on the object side.
Answer: v=−20 cm; virtual, upright, twice as tall
Before moving on: can you explain why your method works, as well as give the answer?
P12-LENS-Q2Lens powerFocused practice
Find the power of a converging lens with focal length 25 cm.
Show your working
Need a starting hint?
Power uses focal length in metres.
Still stuck? Reveal the setup
25 cm=0.25 m.
Show worked solution+
1
Convert
Retain the positive focal sign.
2
Take reciprocal
P=1/0.25=4 dioptres.
Answer: +4 D
Before moving on: can you explain why your method works, as well as give the answer?
P12-LENS-Q3Diverging lensFocused practice
A real object lies 30 cm left of a diverging lens with f=−15 cm. Find v and m.
Show your working
Need a starting hint?
Both u and f are negative.
Still stuck? Reveal the setup
1/v=−1/15−1/30.
Show worked solution+
1
Image position
The reciprocal is −1/10, so v=−10 cm.
2
Magnification
m=(−10)/(−30)=+1/3.
Answer: v=−10 cm; m=+1/3, virtual and upright
Before moving on: can you explain why your method works, as well as give the answer?
Catch a likely mistake
A tempting claim: “Put every stated distance into the formula as positive.”
The formula uses directed distances, not just lengths.
Close the examples and try again
These changed questions test whether you can reconstruct the method. If you use a hint, record where you got stuck and retry later. Completing this small set does not establish full chapter mastery.
P12-LENS-R1Changed distancesIndependent retest
An object is 30 cm left of a converging lens with f=+10 cm. Find v and m without the worked sketch.
Show your working
Need a starting hint?
Use u=−30 cm.
Still stuck? Reveal the setup
1/v=1/10−1/30.
Show worked solution+
1
Image distance
1/v=1/15, so v=15 cm.
2
Magnification
15/(−30)=−0.5.
Answer: v=+15 cm, m=−0.5
Could you solve this without the earlier example? If not, revisit the first line where you got stuck.
P12-LENS-R2Negative powerIndependent retest
A lens has focal length −50 cm. Find power and identify its type in air.
Show your working
Need a starting hint?
Convert to −0.50 m.
Still stuck? Reveal the setup
P=1/(−0.50).
Show worked solution+
1
Compute
The power is −2 D.
2
Interpret
A negative focal length identifies a diverging lens in this setting.
Answer: −2 D; diverging
Could you solve this without the earlier example? If not, revisit the first line where you got stuck.
Choose the next useful step
If a retest exposed the same error, rewrite the first incorrect step and explain its correction aloud. If both were independent, return to a mixed exercise or a missed paper question. A parent can ask what changed in the method rather than only asking for the answer.
What happens when an object is at the focal point?+
In the ideal thin-lens model, the emerging rays are parallel and the image is at infinity. The equation gives 1/v=0; this is not an image at v=0.
Can I print the questions and worked solutions separately?+
Yes. Use the two print buttons above. The question sheet leaves working space and hides solutions; the worked-solutions option includes the solution steps. Your browser can save either view as a PDF.
PERSONAL SUPPORT
Does the same mistake keep returning?
Bring one attempted question and the step that caused difficulty. Discuss suitable TutorMax support for the learner's current level.
