📘 Class 10 Physics – Lenses Numerical Practice
Set 5 – Mixed Problems with HOTS & Assertion-Reason Style
Topic: Light – Refraction through Lenses | CBSE Board Exam Preparation
📝 Instructions
- All questions follow the New Cartesian Sign Convention.
- Distances measured in the direction of incident light (right) are positive.
- Distances measured against the direction of incident light (left) are negative.
- Heights measured upward from the principal axis are positive; downward are negative.
- Use the Lens Formula: \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
- Use the Magnification Formula: \(m=\frac{v}{u}=\frac{h’}{h}\)
- Use the Power Formula: \(P=\frac{1}{f\;(\text{in metres})}\)
- This set includes HOTS, multi-concept, and assertion-reasoning style numericals.
- Click “Show Solution” under each question to view the detailed answer.
🔖 New Cartesian Sign Convention – Quick Reference
| Quantity |
Convex Lens |
Concave Lens |
| Object distance (u) |
Always negative (left of lens) |
Always negative (left of lens) |
| Focal length (f) |
Positive (+) |
Negative (−) |
| Image distance (v) – Real image |
Positive (right side) |
— (not possible for real objects) |
| Image distance (v) – Virtual image |
Negative (same side as object) |
Negative (same side as object) |
Question 1 HOTS – Multi-Concept
A convex lens of power +10 D forms a real image of an object placed 15 cm from the lens. The image so formed acts as a virtual object for a concave lens of power −5 D placed 40 cm from the convex lens (on the same side as the image). Find:
(a) the focal length of each lens,
(b) the image distance from the convex lens,
(c) the position of the virtual object for the concave lens,
(d) the final image distance from the concave lens, and
(e) the total magnification of the system.
Hint: First solve for the convex lens independently (P₁ = +10 D → f₁ = +10 cm, u₁ = −15 cm). The image formed by the convex lens becomes the object for the concave lens. Since this image falls beyond the concave lens, it acts as a virtual object for the concave lens — so u₂ will be positive. This is a rare and tricky sign convention scenario!
Show Solution
Given:
P₁ = +10 D (convex), P₂ = −5 D (concave)
Object distance from convex lens: u₁ = −15 cm
Separation between lenses: d = 40 cm
(a) Focal lengths:
\[
f_1 = \frac{1}{P_1} = \frac{1}{10} = 0.10 \text{ m} = +10 \text{ cm (convex)}
\]
\[
f_2 = \frac{1}{P_2} = \frac{1}{-5} = -0.20 \text{ m} = -20 \text{ cm (concave)}
\]
(b) Image by convex lens:
\[
\frac{1}{f_1} = \frac{1}{v_1} – \frac{1}{u_1}
\]
\[
\frac{1}{10} = \frac{1}{v_1} – \frac{1}{-15} = \frac{1}{v_1} + \frac{1}{15}
\]
\[
\frac{1}{v_1} = \frac{1}{10} – \frac{1}{15} = \frac{3-2}{30} = \frac{1}{30}
\]
\[
v_1 = +30 \text{ cm}
\]
Image by convex lens is at +30 cm (real, on the right side). ✓
(c) Virtual object for concave lens:
The concave lens is at 40 cm from the convex lens.
The image from the convex lens is at 30 cm from the convex lens.
This image is 40 − 30 = 10 cm to the left of the concave lens.
Wait — the image at +30 cm is before the concave lens at +40 cm. So the light converging towards +30 cm gets intercepted by the concave lens at +40 cm. The rays were converging to a point that is now on the right (behind) the concave lens? Let us re-check:
Actually, the convex lens image at +30 cm from the convex lens falls 10 cm before the concave lens (which is at 40 cm). So light has already converged and started diverging again.
Re-reading the problem: The image at 30 cm from the convex lens falls before the concave lens. But the problem says the image “acts as a virtual object.” This means the image at 30 cm is between the two lenses, i.e., 30 cm from the convex = (40 − 30) = 10 cm to the left of the concave lens. After converging at 30 cm, light diverges again and hits the concave lens. So the object for the concave lens is a real object at 10 cm to its left.
For the concave lens:
u₂ = −10 cm (real object, on the left)
f₂ = −20 cm
\[
\frac{1}{f_2} = \frac{1}{v_2} – \frac{1}{u_2}
\]
\[
\frac{1}{-20} = \frac{1}{v_2} – \frac{1}{-10} = \frac{1}{v_2} + \frac{1}{10}
\]
\[
\frac{1}{v_2} = \frac{1}{-20} – \frac{1}{10} = \frac{-1-2}{20} = \frac{-3}{20}
\]
\[
v_2 = -\frac{20}{3} = -6.67 \text{ cm}
\]
(d) Final image:
v₂ = −6.67 cm from the concave lens (negative → same side as object → virtual).
From the convex lens: 40 − 6.67 = 33.33 cm to the right.
(e) Total magnification:
\[
m_1 = \frac{v_1}{u_1} = \frac{30}{-15} = -2
\]
\[
m_2 = \frac{v_2}{u_2} = \frac{-20/3}{-10} = +\frac{2}{3}
\]
\[
m_{\text{total}} = m_1 \times m_2 = (-2) \times \left(+\frac{2}{3}\right) = -\frac{4}{3} \approx -1.33
\]
Total magnification = −4/3 ≈ −1.33
The final image is real (formed by convergence of refracted rays from convex lens, then made virtual by concave lens), inverted, and slightly magnified.
Question 2 Assertion-Reason Style
Assertion (A): A convex lens of focal length 20 cm always produces a magnified image when the object is placed at 25 cm from the lens.
Reason (R): When the object is placed between F and 2F of a convex lens, the image formed is real, inverted, and magnified.
(a) Verify the assertion numerically by finding v, m, and h′ if h = 3 cm.
(b) Is the Reason a correct explanation of the Assertion?
(c) Find the power of this lens.
Hint: f = +20 cm. 2F = 40 cm. Object at 25 cm → between F (20 cm) and 2F (40 cm). u = −25 cm. Apply the lens formula. Check if |m| > 1 to verify “magnified.” Then reason whether the assertion and reason are both true and if R explains A.
Show Solution
Given:
f = +20 cm (convex lens), u = −25 cm, h = +3 cm
F = 20 cm, 2F = 40 cm
Since 20 < 25 < 40, object is between F and 2F. ✓
(a) Numerical verification:
\[
\frac{1}{f} = \frac{1}{v} – \frac{1}{u}
\]
\[
\frac{1}{20} = \frac{1}{v} – \frac{1}{-25} = \frac{1}{v} + \frac{1}{25}
\]
\[
\frac{1}{v} = \frac{1}{20} – \frac{1}{25} = \frac{5-4}{100} = \frac{1}{100}
\]
\[
v = +100 \text{ cm}
\]
Magnification:
\[
m = \frac{v}{u} = \frac{100}{-25} = -4
\]
Image height:
\[
h’ = m \times h = (-4) \times 3 = -12 \text{ cm}
\]
|m| = 4 > 1 → Image is magnified. ✓
m is negative → Image is inverted. ✓
v is positive → Image is real. ✓
Image height = 12 cm (inverted, below principal axis).
The Assertion is TRUE. ✓
(b) Evaluation of Reason:
The Reason states: “When the object is between F and 2F, the image is real, inverted, and magnified.”
This is a standard rule from ray optics and is TRUE. ✓
Since the object at 25 cm is indeed between F (20 cm) and 2F (40 cm), the Reason correctly explains why the Assertion is true.
Both A and R are true, and R is the correct explanation of A.
(c) Power:
\[
P = \frac{1}{f \text{ (in m)}} = \frac{1}{0.20} = +5 \text{ D}
\]
Power = +5 D
Question 3 HOTS – Reverse Engineering
A lens forms an image of height −6 cm (inverted) of an object of height 2 cm. The image is formed at 45 cm on the other side of the lens. Determine:
(a) the type of lens,
(b) the magnification,
(c) the object distance,
(d) the focal length, and
(e) the power of the lens.
All quantities must be derived — no lens type is given directly.
Hint: Image is inverted (h′ = −6 cm) and on the other side → real image → v = +45 cm. Only a convex lens can form a real, inverted image of a real object. Find m from h′/h, then u from m = v/u. Then apply the lens formula for f. This is a complete reverse-engineering problem.
Show Solution
Given:
h = +2 cm, h′ = −6 cm (inverted), v = +45 cm (other side → real image)
(a) Type of lens:
The image is real and inverted. Only a convex lens can form a real, inverted image of a real object.
∴ The lens is a convex lens. ✓
(b) Magnification:
\[
m = \frac{h’}{h} = \frac{-6}{2} = -3
\]
Magnification = −3 (negative → inverted, |m| > 1 → magnified). ✓
(c) Object distance:
\[
m = \frac{v}{u} \implies -3 = \frac{45}{u} \implies u = \frac{45}{-3} = -15 \text{ cm}
\]
Object distance = −15 cm. ✓
(d) Focal length:
\[
\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{45} – \frac{1}{-15} = \frac{1}{45} + \frac{1}{15}
\]
\[
\frac{1}{f} = \frac{1}{45} + \frac{3}{45} = \frac{4}{45}
\]
\[
f = +\frac{45}{4} = +11.25 \text{ cm}
\]
Focal length = +11.25 cm (positive → confirms convex lens). ✓
Verification: Object at 15 cm, F = 11.25 cm, 2F = 22.5 cm.
Object is between F and 2F (11.25 < 15 < 22.5) → image should be real, inverted, magnified. ✓ Consistent!
(e) Power:
\[
P = \frac{1}{f \text{ (in m)}} = \frac{1}{0.1125} = +8.89 \text{ D}
\]
Power ≈ +8.89 D
Question 4 HOTS – Image on Same Side Trap
An object is placed 8 cm from a convex lens of focal length 12 cm. Find:
(a) the image distance,
(b) the magnification,
(c) the nature of the image (real or virtual, erect or inverted, magnified or diminished),
(d) the power of the lens, and
(e) if the object height is 4 cm, find the image height.
A student says the image must be real because a convex lens is a converging lens. Is the student correct? Justify.
Hint: f = +12 cm, u = −8 cm. Notice that |u| = 8 < f = 12, so the object is between the optical centre and the focus. This is the magnifying glass case! Expect v to be negative (virtual image). The student’s claim is a classic misconception trap.
Show Solution
Given:
f = +12 cm (convex lens), u = −8 cm, h = +4 cm
Note: |u| = 8 cm < f = 12 cm → object is between O and F.
(a) Image distance:
\[
\frac{1}{f} = \frac{1}{v} – \frac{1}{u}
\]
\[
\frac{1}{12} = \frac{1}{v} – \frac{1}{-8} = \frac{1}{v} + \frac{1}{8}
\]
\[
\frac{1}{v} = \frac{1}{12} – \frac{1}{8} = \frac{2-3}{24} = \frac{-1}{24}
\]
\[
v = -24 \text{ cm}
\]
Image distance = −24 cm (negative → image on the same side as object → virtual). ✓
(b) Magnification:
\[
m = \frac{v}{u} = \frac{-24}{-8} = +3
\]
Magnification = +3 (positive → erect; |m| = 3 > 1 → magnified). ✓
(c) Nature of image:
• v is negative → Virtual
• m is positive → Erect
• |m| > 1 → Magnified
The image is virtual, erect, and magnified.
(d) Power:
\[
P = \frac{1}{f \text{ (in m)}} = \frac{1}{0.12} = +8.33 \text{ D}
\]
Power ≈ +8.33 D
(e) Image height:
\[
h’ = m \times h = (+3) \times 4 = +12 \text{ cm}
\]
Image height = +12 cm (positive → erect, above principal axis). ✓
Is the student correct?
No. The student is incorrect. Although a convex lens is a converging lens, it does NOT always form a real image.
When the object is placed between the optical centre (O) and the principal focus (F) — i.e., when |u| < f — the convex lens acts as a magnifying glass and forms a virtual, erect, and magnified image on the same side as the object.
Key Rule:
• Object beyond F → Real image
• Object at F → Image at infinity
• Object between O and F → Virtual image
Question 5 HOTS – Equal Size Image
A convex lens produces a real, inverted image of the same size as the object. The distance between the object and the image is 80 cm. Find:
(a) the object distance,
(b) the image distance,
(c) the focal length of the lens,
(d) the power of the lens, and
(e) where exactly (in terms of F and 2F) is the object placed?
Hint: Same size, real, inverted → m = −1. This means |v| = |u|. Object and image are on opposite sides, so the total distance = |u| + |v| = 80 cm. Since |v| = |u|, each is 40 cm. A same-size real image is formed when the object is at 2F. Therefore f = |u|/2.
Show Solution
Given:
m = −1 (real, inverted, same size)
Distance between object and image = 80 cm
Key Concept: When m = −1, the image is at the same distance from the lens as the object but on the opposite side.
\[
m = \frac{v}{u} = -1 \implies v = -u
\]
Since u is negative (say u = −x), we get v = +x.
(a) & (b) Object and Image distance:
Distance between object and image:
\[
|u| + |v| = x + x = 2x = 80 \text{ cm}
\]
\[
x = 40 \text{ cm}
\]
\[
u = -40 \text{ cm}, \quad v = +40 \text{ cm}
\]
Object distance = −40 cm, Image distance = +40 cm. ✓
(c) Focal length:
\[
\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{40} – \frac{1}{-40} = \frac{1}{40} + \frac{1}{40} = \frac{2}{40} = \frac{1}{20}
\]
\[
f = +20 \text{ cm}
\]
Focal length = +20 cm. ✓
(d) Power:
\[
P = \frac{1}{f \text{ (in m)}} = \frac{1}{0.20} = +5 \text{ D}
\]
Power = +5 D. ✓
(e) Position of object:
\[
2F = 2 \times 20 = 40 \text{ cm}
\]
|u| = 40 cm = 2F.
The object is placed at 2F (the centre of curvature).
Rule: When an object is placed at 2F of a convex lens, the image is formed at 2F on the other side, and is real, inverted, and of the same size.
Question 6 HOTS – Comparing Two Lenses
Two lenses, Lens A (convex, f = 15 cm) and Lens B (concave, f = 15 cm), are each placed 30 cm from an object of height 6 cm. For each lens, find:
(a) the image distance,
(b) the magnification,
(c) the image height, and
(d) the nature of the image.
Then: (e) Compare the two results and explain why a convex and concave lens of the same focal length magnitude produce very different images.
Hint: For Lens A: f = +15 cm. For Lens B: f = −15 cm. Both have u = −30 cm. Apply the lens formula separately. The comparison will reveal how the sign of f completely changes the image.
Show Solution
Given:
Lens A: f_A = +15 cm (convex), Lens B: f_B = −15 cm (concave)
u = −30 cm (both cases), h = +6 cm
LENS A (Convex, f = +15 cm):
(a) Image distance:
\[
\frac{1}{f_A} = \frac{1}{v_A} – \frac{1}{u}
\]
\[
\frac{1}{15} = \frac{1}{v_A} + \frac{1}{30}
\]
\[
\frac{1}{v_A} = \frac{1}{15} – \frac{1}{30} = \frac{2-1}{30} = \frac{1}{30}
\]
\[
v_A = +30 \text{ cm}
\]
(b) Magnification:
\[
m_A = \frac{v_A}{u} = \frac{30}{-30} = -1
\]
(c) Image height:
\[
h’_A = m_A \times h = (-1) \times 6 = -6 \text{ cm}
\]
(d) Nature: v positive →
Real; m = −1 →
Inverted; |m| = 1 →
Same size.
Image:
Real, inverted, same size (at 2F). ✓
LENS B (Concave, f = −15 cm):
(a) Image distance:
\[
\frac{1}{f_B} = \frac{1}{v_B} – \frac{1}{u}
\]
\[
\frac{1}{-15} = \frac{1}{v_B} + \frac{1}{30}
\]
\[
\frac{1}{v_B} = \frac{1}{-15} – \frac{1}{30} = \frac{-2-1}{30} = \frac{-3}{30} = \frac{-1}{10}
\]
\[
v_B = -10 \text{ cm}
\]
(b) Magnification:
\[
m_B = \frac{v_B}{u} = \frac{-10}{-30} = +\frac{1}{3}
\]
(c) Image height:
\[
h’_B = m_B \times h = \left(+\frac{1}{3}\right) \times 6 = +2 \text{ cm}
\]
(d) Nature: v negative →
Virtual; m positive →
Erect; |m| = 1/3 < 1 →
Diminished.
Image:
Virtual, erect, diminished. ✓
(e) Comparison:
| Property |
Lens A (Convex) |
Lens B (Concave) |
| v |
+30 cm |
−10 cm |
| m |
−1 |
+1/3 |
| h′ |
−6 cm (inverted) |
+2 cm (erect) |
| Nature |
Real, inverted, same size |
Virtual, erect, diminished |
Explanation: Even though both lenses have the same magnitude of focal length (15 cm), the
sign of f makes all the difference. A convex lens (f > 0)
converges light to form a real image, while a concave lens (f < 0)
diverges light, always forming a virtual image. The lens formula’s behaviour changes fundamentally based on the sign of f. This is why the
sign convention is critical — the same number with a different sign produces a completely different physical outcome.
Question 7 HOTS – Unknown Lens Identification
A lens forms an image at −12 cm when an object is placed at 36 cm from it. The image is erect and diminished. Without being told the lens type, determine:
(a) the sign-convention-corrected values of u and v,
(b) the focal length of the lens,
(c) the type of lens,
(d) the magnification,
(e) the power of the lens, and
(f) the image height if the object height is 9 cm.
Hint: Object is always on the left → u = −36 cm. Image at −12 cm → same side as object → virtual. Erect + virtual + diminished → think about which lens always does this. Apply the lens formula. Check whether f is positive or negative to identify the lens.
Show Solution
Given:
Image at −12 cm, object at 36 cm from lens, image is erect and diminished.
(a) Sign-convention-corrected values:
u = −36 cm (object on the left)
v = −12 cm (image on the same side as object → virtual)
(b) Focal length:
\[
\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-12} – \frac{1}{-36} = \frac{-1}{12} + \frac{1}{36}
\]
\[
\frac{1}{f} = \frac{-3 + 1}{36} = \frac{-2}{36} = \frac{-1}{18}
\]
\[
f = -18 \text{ cm}
\]
Focal length = −18 cm. ✓
(c) Type of lens:
f is negative → The lens is a concave (diverging) lens. ✓
Cross-check: Concave lens always forms virtual, erect, diminished images → matches the given data. ✓
(d) Magnification:
\[
m = \frac{v}{u} = \frac{-12}{-36} = +\frac{1}{3}
\]
Magnification = +1/3 (positive → erect; |m| < 1 → diminished). ✓
(e) Power:
\[
P = \frac{1}{f \text{ (in m)}} = \frac{1}{-0.18} = -5.56 \text{ D}
\]
Power ≈ −5.56 D (negative → diverging lens). ✓
(f) Image height:
\[
h’ = m \times h = \left(+\frac{1}{3}\right) \times 9 = +3 \text{ cm}
\]
Image height = +3 cm (positive → erect, above principal axis). ✓
Question 8 HOTS – Object at 2F Challenge
An object of height 5 cm is placed at a distance equal to twice the focal length of a convex lens. The focal length of the lens is 18 cm.
(a) Find the object distance (with sign).
(b) Find the image distance.
(c) Find the magnification and image height.
(d) Without calculation, predict: If the object is now shifted 1 cm closer to the lens (from 2F), will the image move closer to or farther from the lens? Will it become larger or smaller? Verify your prediction numerically.
Hint: f = +18 cm, 2F = 36 cm → u = −36 cm. At 2F, the image forms at 2F on the other side (m = −1). For part (d), moving the object from 2F towards F means the object enters the “between F and 2F” zone — recall what happens to the image in this zone. Verify using u = −35 cm.
Show Solution
Given:
f = +18 cm, h = +5 cm
Object at 2F → |u| = 2f = 36 cm → u = −36 cm
(a) Object distance:
u = −36 cm. ✓
(b) Image distance:
\[
\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{v} + \frac{1}{36}
\]
\[
\frac{1}{18} = \frac{1}{v} + \frac{1}{36}
\]
\[
\frac{1}{v} = \frac{1}{18} – \frac{1}{36} = \frac{2-1}{36} = \frac{1}{36}
\]
\[
v = +36 \text{ cm}
\]
Image distance = +36 cm (at 2F on the other side). ✓
(c) Magnification and image height:
\[
m = \frac{v}{u} = \frac{36}{-36} = -1
\]
\[
h’ = m \times h = (-1) \times 5 = -5 \text{ cm}
\]
Magnification = −1, Image height = −5 cm (inverted, same size). ✓
(d) Prediction (object moved 1 cm closer, to u = −35 cm):
Prediction: Moving from 2F towards F places the object in the “between F and 2F” region. In this region, the image moves farther away (beyond 2F) and becomes larger (|m| > 1).
Numerical Verification (u = −35 cm, f = +18 cm):
\[
\frac{1}{18} = \frac{1}{v} + \frac{1}{35}
\]
\[
\frac{1}{v} = \frac{1}{18} – \frac{1}{35} = \frac{35 – 18}{630} = \frac{17}{630}
\]
\[
v = \frac{630}{17} = +37.06 \text{ cm}
\]
\[
m = \frac{37.06}{-35} = -1.059
\]
Results:
• v increased from 36 cm → 37.06 cm → image moved farther. ✅
• |m| increased from 1 → 1.059 → image became larger. ✅
Prediction confirmed! Even a tiny 1 cm shift from 2F towards F causes the image to shift farther and grow larger. As the object approaches F, v → ∞ and |m| → ∞.
Question 9 HOTS – Concave Lens Maximum Image Distance
A concave lens has a focal length of 25 cm. An object is moved from infinity towards the lens and stops at 10 cm from the lens. Calculate the image distance for the following object positions:
(a) u = ∞ (very far away)
(b) u = −100 cm
(c) u = −50 cm
(d) u = −25 cm (at F)
(e) u = −10 cm
Then: (f) What is the range within which the image always lies for a concave lens? What is the maximum possible image distance from the lens?
Hint: f = −25 cm. For each value of u, apply the lens formula. As u goes from −∞ to 0, watch how v changes. The image for a concave lens is always between the optical centre and the focus. This is a pattern-recognition HOTS question.
Show Solution
Given: f = −25 cm (concave lens)
Using the lens formula: \(\frac{1}{f} = \frac{1}{v} – \frac{1}{u}\)
(a) u = ∞ (object at infinity):
\[
\frac{1}{-25} = \frac{1}{v} – 0 \implies v = -25 \text{ cm}
\]
v =
−25 cm (image at F, virtual). ✓
(b) u = −100 cm:
\[
\frac{1}{-25} = \frac{1}{v} + \frac{1}{100}
\]
\[
\frac{1}{v} = \frac{-1}{25} – \frac{1}{100} = \frac{-4-1}{100} = \frac{-5}{100} = \frac{-1}{20}
\]
\[
v = -20 \text{ cm}
\]
v =
−20 cm. ✓
(c) u = −50 cm:
\[
\frac{1}{-25} = \frac{1}{v} + \frac{1}{50}
\]
\[
\frac{1}{v} = \frac{-1}{25} – \frac{1}{50} = \frac{-2-1}{50} = \frac{-3}{50}
\]
\[
v = \frac{-50}{3} = -16.67 \text{ cm}
\]
v =
−16.67 cm. ✓
(d) u = −25 cm (object at F):
\[
\frac{1}{-25} = \frac{1}{v} + \frac{1}{25}
\]
\[
\frac{1}{v} = \frac{-1}{25} – \frac{1}{25} = \frac{-2}{25}
\]
\[
v = \frac{-25}{2} = -12.5 \text{ cm}
\]
v =
−12.5 cm. ✓
(e) u = −10 cm:
\[
\frac{1}{-25} = \frac{1}{v} + \frac{1}{10}
\]
\[
\frac{1}{v} = \frac{-1}{25} – \frac{1}{10} = \frac{-2-5}{50} = \frac{-7}{50}
\]
\[
v = \frac{-50}{7} = -7.14 \text{ cm}
\]
v =
−7.14 cm. ✓
(f) Pattern & Range:
| |u| (cm) |
v (cm) |
|v| (cm) |
| ∞ | −25 | 25 (= |f|) |
| 100 | −20 | 20 |
| 50 | −16.67 | 16.67 |
| 25 (= |f|) | −12.5 | 12.5 (= |f|/2) |
| 10 | −7.14 | 7.14 |
Observation:
• As the object moves from infinity towards the lens, |v|
decreases from |f| towards 0.
• The image is
always between the
optical centre (O) and the
focus (F) of the concave lens.
Range of image: The image distance for a concave lens always satisfies:
\[
0 < |v| \leq |f|
\]
Maximum image distance = |f| = 25 cm (occurs when the object is at infinity).
The image can
never be farther from the lens than the focal point.
Question 10 HOTS – Board Exam Multi-Part Mega Question
A student performs an experiment with a convex lens of focal length 15 cm and records the following three cases:
Case I: Object placed at 45 cm from the lens.
Case II: Object placed at 30 cm from the lens.
Case III: Object placed at 10 cm from the lens.
For each case, find:
(a) the image distance,
(b) the magnification, and
(c) the nature of the image.
Then:
(d) In which case is the image the largest?
(e) In which case is the image virtual?
(f) In which case does the object lie at 2F?
(g) Rank all three cases in increasing order of image size.
Hint: f = +15 cm, so F = 15 cm and 2F = 30 cm. Case I: object beyond 2F (|u| > 2f). Case II: object at 2F (|u| = 2f). Case III: object between O and F (|u| < f). Each case produces a distinctly different image. This is the ultimate position-analysis question.
Show Solution
Given:
f = +15 cm (convex lens), F = 15 cm, 2F = 30 cm
CASE I: u = −45 cm (Object beyond 2F)
Since |u| = 45 > 2F = 30 →
Object is beyond 2F.
(a) Image distance:
\[
\frac{1}{15} = \frac{1}{v} + \frac{1}{45}
\]
\[
\frac{1}{v} = \frac{1}{15} – \frac{1}{45} = \frac{3-1}{45} = \frac{2}{45}
\]
\[
v = +22.5 \text{ cm}
\]
(b) Magnification:
\[
m = \frac{v}{u} = \frac{22.5}{-45} = -0.5
\]
(c) Nature: v > 0 → Real; m < 0 → Inverted; |m| = 0.5 < 1 → Diminished.
Image:
Real, inverted, diminished. ✓
Image lies between F and 2F on the other side (15 < 22.5 < 30). ✓
CASE II: u = −30 cm (Object at 2F)
Since |u| = 30 = 2F →
Object is at 2F.
(a) Image distance:
\[
\frac{1}{15} = \frac{1}{v} + \frac{1}{30}
\]
\[
\frac{1}{v} = \frac{1}{15} – \frac{1}{30} = \frac{2-1}{30} = \frac{1}{30}
\]
\[
v = +30 \text{ cm}
\]
(b) Magnification:
\[
m = \frac{30}{-30} = -1
\]
(c) Nature: Real, inverted,
same size. ✓
Image at 2F on the other side. ✓
CASE III: u = −10 cm (Object between O and F)
Since |u| = 10 < F = 15 →
Object is between O and F.
(a) Image distance:
\[
\frac{1}{15} = \frac{1}{v} + \frac{1}{10}
\]
\[
\frac{1}{v} = \frac{1}{15} – \frac{1}{10} = \frac{2-3}{30} = \frac{-1}{30}
\]
\[
v = -30 \text{ cm}
\]
(b) Magnification:
\[
m = \frac{-30}{-10} = +3
\]
(c) Nature: v < 0 → Virtual; m > 0 → Erect; |m| = 3 > 1 → Magnified.
Image:
Virtual, erect, magnified. ✓
COMPARISON & ANSWERS:
| Property |
Case I (u = −45) |
Case II (u = −30) |
Case III (u = −10) |
| Position |
Beyond 2F |
At 2F |
Between O and F |
| v (cm) |
+22.5 |
+30 |
−30 |
| m |
−0.5 |
−1 |
+3 |
| |m| |
0.5 |
1 |
3 |
| Nature |
Real, inverted, diminished |
Real, inverted, same size |
Virtual, erect, magnified |
(d) Largest image: Case III (|m| = 3, the largest). ✓
(e) Virtual image: Case III (v is negative, object between O and F). ✓
(f) Object at 2F: Case II (|u| = 30 = 2F). ✓
(g) Increasing order of image size (|m|):
\[
\text{Case I } (|m|=0.5) \;<\; \text{Case II } (|m|=1) \;<\; \text{Case III } (|m|=3)
\]
Case I < Case II < Case III
Key Takeaway: As the object moves from beyond 2F towards the lens (towards F and then between O and F), the image size
continuously increases. The transition from real to virtual happens when the object crosses the focal point.
📊 Set 5 – Master Summary Table
| Q# |
Type |
Lens |
u (cm) |
f (cm) |
v (cm) |
m |
Key Concept |
| 1 |
HOTS |
Convex + Concave |
−15 |
+10, −20 |
+30 → −6.67 |
−4/3 |
Two-lens system, virtual object |
| 2 |
Assertion-Reason |
Convex |
−25 |
+20 |
+100 |
−4 |
Between F & 2F verification |
| 3 |
HOTS |
Convex |
−15 |
+11.25 |
+45 |
−3 |
Full reverse engineering |
| 4 |
HOTS |
Convex |
−8 |
+12 |
−24 |
+3 |
Convex lens virtual image trap |
| 5 |
HOTS |
Convex |
−40 |
+20 |
+40 |
−1 |
Same-size image at 2F |
| 6 |
HOTS |
Convex vs Concave |
−30 |
±15 |
+30 / −10 |
−1 / +1/3 |
Same |f|, opposite lens comparison |
| 7 |
HOTS |
Unknown → Concave |
−36 |
−18 |
−12 |
+1/3 |
Identify lens from image data |
| 8 |
HOTS |
Convex |
−36 → −35 |
+18 |
+36 → +37.06 |
−1 → −1.06 |
2F prediction + shift verification |
| 9 |
HOTS |
Concave |
∞ to −10 |
−25 |
−25 to −7.14 |
varies |
Image range & maximum v |
| 10 |
Mega HOTS |
Convex (3 cases) |
−45, −30, −10 |
+15 |
+22.5, +30, −30 |
−0.5, −1, +3 |
Complete position analysis & ranking |
🧠 Key Concepts Tested in Set 5
| Concept |
What to Remember |
| Two-Lens System |
Image of first lens = object for second lens. Recalculate u₂ based on geometry. Total m = m₁ × m₂. |
| Convex Lens Virtual Image |
When object is between O and F (|u| < f), convex lens gives virtual, erect, magnified image. v is negative. |
| Object at 2F |
Image at 2F on other side. m = −1. Same size, real, inverted. |
| Concave Lens Image Range |
Image always between O and F. Maximum |v| = |f| (when object at ∞). Always virtual, erect, diminished. |
| Reverse Engineering |
Given h, h′, v → find lens type, u, f, P. Use m = h′/h first, then m = v/u, then lens formula. |
| Position Shifting |
Moving object from 2F toward F → image goes farther, becomes larger. At F → v = ∞. |
📘 Class 10 Physics – Lenses Numerical Practice | Set 5 – Mixed Problems with HOTS & Assertion-Reason
Prepared for CBSE Board Exam Practice | New Cartesian Sign Convention
Previous: Set 4 – Board-Level Tricky Problems | Next: Set 6 – Full-Length Mock Test
