NEB Class 11 • Physics • Worked Practice

Lenses Practice Set: NEB 11 Physics

Solve a graduated set where every answer must agree with a diagram, a sign convention and image language. Use the worked reasoning to mark decisions, not just numbers.

  • Retrieval, ray and calculation rounds
  • Fully worked examples with physical checks
  • Practical analysis and timed mixed test
Convex lens ray diagramTwo principal rays from an object pass through a convex lens and meet at a real inverted image.
A diagram should predict the sign and size of the image.

Source and prerequisite check

Keep practice inside a defensible model

The CDC secondary curriculum and CDC Physics Grade 11 page define the Nepal learning context. Review the Lenses concept guide and Lenses Study Guide before the challenge round. The equations and ray rules align with OpenStax lens image formation.

Unless stated otherwise, use the common real-is-positive convention described in the source: do positive for a real object, di positive for a real image, f positive for a converging lens and negative for a diverging lens.

Round 1 • Retrieval

Concept questions before arithmetic

  1. Define optical centre, principal axis, principal focus and focal length.
  2. State three principal-ray rules for a convex lens.
  3. Describe the image as a real object moves from far beyond 2F to inside F.
  4. Explain why a concave lens gives a virtual image for a real object.
  5. Distinguish real and virtual images using the screen test.
  6. Explain why covering half a lens reduces brightness but does not neatly remove half the image.
  7. State the SI unit of lens power and the focal-length unit required.
  8. Name two assumptions behind thin-lens equations.

Answer framework

A strong response names the ray behaviour, says whether actual rays meet or only backward extensions meet, and connects the result to a screen. For assumptions, mention a thin element, paraxial rays, small aperture relative to curvature, negligible aberration and a uniform surrounding medium as appropriate.

Round 2 • Constructions

Six ray-diagram tasks

  1. Convex lens, object beyond 2F.
  2. Convex lens, object at 2F.
  3. Convex lens, object between F and 2F.
  4. Convex lens, object at F; explain the limiting result.
  5. Convex lens, object inside F.
  6. Concave lens with a real object at an arbitrary distance.

For each diagram, award one mark each for equal focal marks, correct object region, two correct principal rays, arrowheads, correct solid/dashed lines, accurate image placement and complete image description.

Reasoning check

In task 3, the parallel ray refracts through the far focus and the optical-centre ray continues straight. They meet beyond 2F, so the image is real, inverted and enlarged. If your drawing places it between F and 2F, inspect the object region or focal point before changing labels.

Round 3 • Worked calculations

Change the optical case, not only the numbers

1. Convex lens beyond 2F

f=10 cm, do=30 cm. 1/di=1/10−1/30=1/15, so di=15 cm and m=−0.50. The image is between F and 2F, real, inverted and diminished.

2. Object between F and 2F

f=12 cm, do=18 cm. 1/di=1/12−1/18=1/36, so di=36 cm and m=−2. The image is beyond 2F, real, inverted and enlarged.

3. Convex magnifier

f=15 cm, do=10 cm. 1/di=1/15−1/10=−1/30, so di=−30 cm and m=+3. The virtual image lies on the object side, upright and enlarged.

4. Diverging lens

f=−20 cm, do=40 cm. 1/di=−1/20−1/40=−3/40, so di=−13.3 cm and m≈+0.333. The virtual image is between O and F, upright and diminished.

5. Find the object

A +10 cm lens forms a real image 30 cm away. 1/do=1/10−1/30=1/15, so do=15 cm. The object is between F and 2F and the image magnification is −2.

6. Find focal length

A real object at 24 cm gives a real image at 48 cm. 1/f=1/24+1/48=1/16, so f=16 cm. The object lies between F and 2F, matching an enlarged real image.

Round 4 • Power

Reciprocal and combination problems

GivenWorkingAnswer and meaning
f=+50 cmf=+0.50 m; P=1/f+2.0 D, converging
P=−5 Df=1/P−0.20 m, diverging
+6 D and −2 D in contactPeq=4 Df=+0.25 m
f=+25 cm and +50 cmP=4+2+6 D, f≈0.167 m

Extension: design two lens powers in contact that produce +3 D, then give two different valid pairs. Explain why many combinations share the same paraxial equivalent power but may differ in thickness, aberration and construction.

Round 5 • Practical reasoning

Analyse a focal-length investigation

A student records do=(30, 35, 40, 50) cm and corresponding focused di values. Their lens stand has a 2 cm-wide base, but they measure from its front edge. Identify the systematic reference error and propose measuring from the marked optical-centre plane instead.

  1. Explain why all object distances for a screen image must exceed f.
  2. Describe alignment checks before recording.
  3. Explain how focusing from both directions reduces judgement bias.
  4. State how repeated readings reveal random spread.
  5. Derive the expected line for y=1/di against x=1/do.
  6. Explain why a single distant-object value is only an estimate of f.

The straight-line form is y=−x+1/f. A slope far from −1 may indicate axis definitions, distance references, units or model limitations. Use PhET Geometric Optics to compare ideal movements, but do not copy ideal data as an experiment.

Timed challenge

Thirty-five minute mixed paper

  1. Draw an unknown converging-lens case where m=−1 and infer object and image regions.
  2. A +8 D lens receives an object 20 cm away. Find image distance and magnification.
  3. A −4 D lens receives an object 50 cm away. Describe and calculate the image.
  4. Find the equivalent power of +5 D, +2 D and −3 D thin lenses in contact.
  5. Explain why a camera sensor must occupy a real-image plane.
  6. Critique “a convex lens always magnifies.”
  7. Use a graph intercept 6.25 m⁻¹ to find f and power.
  8. Explain two uncertainties and one safety rule in a lens practical.

Solutions: task 1 corresponds to object and image at 2F. In task 2, f=12.5 cm; 1/di=8−5=3 m⁻¹, so di=0.333 m and m≈−1.67. Task 3 has f=−0.25 m and do=0.50 m; di≈−0.167 m and m≈+0.333. Task 4 gives +4 D. Task 7 gives f=0.160 m and +6.25 D.

Mark the first wrong decision, not just the final line. Retest an isomorphic question after two days. For online or physical NEB tuition, call 9846662070.

Frequently asked questions

Questions about the lenses practice set

Should I look at a solution after one failed attempt?

First make a direction prediction, inspect the diagram and identify the first uncertain step. Then study the worked reasoning and redo the problem from an empty page.

How should I mark a ray diagram?

Score focal marks, object region, ray rules, arrows, solid versus dashed lines, image placement and complete image description separately.

Why are the questions deliberately different?

Changing object region, lens type and unknown tests model selection; replacing only numbers can create false fluency.

What if my school uses another sign convention?

Translate every definition consistently, then use the matching formula. Physical image location and nature must remain unchanged.

What score shows readiness?

Aim for at least 80% with no repeated ray-rule, sign, unit or interpretation error on a fresh mixed set.

Where can I get help with lens problems?

Call 9846662070 for current online or physical NEB tuition options and confirm timetable and fees.

References and next steps

Sources and related study guides

After marking, continue to Dispersion. Curriculum scope and sources were checked on 2 August 2026; follow current CDC, NEB and college instructions if requirements change.

Extension round

Reverse and design problems

Image requirement to object position

A +5 D lens must produce a real image three times the object size. Its focal length is 0.20 m and m=−3, so di=3do. Substitution into 1/f=1/do+1/(3do) gives do=4f/3≈0.267 m and di=0.800 m.

Virtual target image

A convex lens with f=12 cm should give an upright image twice the object height. m=+2 implies di=−2do. The lens equation gives 1/12=1/do−1/(2do)=1/(2do), so do=6 cm and di=−12 cm. The object lies inside F as required.

Now write a third reverse problem where the answer must be a diverging lens. Include enough information to infer both sign and magnitude. Mark whether the solver identifies image nature before algebra and rejects any positive-focal-length answer that contradicts the specified diminished upright image for a real object.

Final check: interpret every sign.

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