Class 10th Lens Question Set-4

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Class 10 Physics – Lenses Numerical Practice (Set 4)

πŸ“˜ Class 10 Physics – Lenses Numerical Practice

Set 4 – Board-Level Tricky Problems

Topic: Light – Refraction through Lenses | CBSE Board Exam Preparation

πŸ“ 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:   \(P=\frac{1}{f\;(\text{in metres})}\)
  • Click “Show Solution” under each question to view the detailed answer.

πŸ”– 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 (βˆ’)
Image distance (v) – Real image Positive (right side) β€” (concave lens does not form real images of real objects)
Image distance (v) – Virtual image Negative (same side as object) Negative (same side as object)

Question 1

An object of height 5 cm is placed at a distance of 30 cm from a convex lens. A real, inverted image of height 10 cm is formed on the other side of the lens. Find:
(a) the magnification produced,
(b) the image distance,
(c) the focal length of the lens, and
(d) the power of the lens.

Hint: Object is always on the left β†’ u is negative. Real image by a convex lens is formed on the right β†’ v is positive. The image is inverted β†’ hβ€² is negative. Focal length of convex lens is positive.

Show Solution
Given:
Object height, h = +5 cm
Object distance, u = βˆ’30 cm
Image height, hβ€² = βˆ’10 cm (real & inverted β†’ negative)

(a) Magnification:
\[ m = \frac{h’}{h} = \frac{-10}{5} = -2 \] The magnification is βˆ’2 (negative confirms inverted image).

(b) Image distance (v):
Using \( m = \frac{v}{u} \):
\[ -2 = \frac{v}{-30} \] \[ v = (-2) \times (-30) = +60 \text{ cm} \] Image distance, v = +60 cm (positive β†’ real image on opposite side). βœ“

(c) Focal length (f):
Using the lens formula:
\[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{60} – \frac{1}{-30} = \frac{1}{60} + \frac{1}{30} \] \[ \frac{1}{f} = \frac{1}{60} + \frac{2}{60} = \frac{3}{60} = \frac{1}{20} \] \[ f = +20 \text{ cm} \] Focal length = +20 cm (positive β†’ convex lens). βœ“

(d) Power of the lens:
\[ P = \frac{1}{f \text{ (in metres)}} = \frac{1}{0.20} = +5 \text{ D} \] Power = +5 D

Question 2

A concave lens of focal length 20 cm forms an image at a distance of 15 cm from the lens on the same side as the object. Find:
(a) the object distance,
(b) the magnification, and
(c) the nature of the image.

Hint: Concave lens β†’ f is negative (f = βˆ’20 cm). Image on the same side as the object β†’ v is negative (v = βˆ’15 cm). Object distance u is always negative.

Show Solution
Given:
f = βˆ’20 cm (concave lens)
v = βˆ’15 cm (image on same side as object)

(a) Object distance (u):
Using the lens formula:
\[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \] \[ \frac{1}{-20} = \frac{1}{-15} – \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{-15} – \frac{1}{-20} = \frac{1}{-15} + \frac{1}{20} \] \[ \frac{1}{u} = \frac{-4 + 3}{60} = \frac{-1}{60} \] \[ u = -60 \text{ cm} \] Object distance = βˆ’60 cm (i.e., 60 cm to the left of the lens). βœ“

(b) Magnification:
\[ m = \frac{v}{u} = \frac{-15}{-60} = +\frac{1}{4} = +0.25 \] Magnification = +0.25

(c) Nature of the image:
β€’ m is positive β†’ image is erect.
β€’ |m| = 0.25 < 1 β†’ image is diminished.
β€’ v is negative β†’ image is on the same side as object β†’ virtual.

The image is virtual, erect, and diminished.

Question 3

A convex lens produces a virtual, erect image that is 3 times the size of the object. If the focal length of the lens is 15 cm, find:
(a) the object distance,
(b) the image distance, and
(c) the power of the lens.

Hint: A convex lens forms a virtual image only when the object is placed between the optical centre and the focus (u < f). Virtual & erect β†’ m = +3. v will be negative (same side as object). f is positive for convex lens. This is the classic “magnifying glass” case β€” a common CBSE trap!

Show Solution
Given:
f = +15 cm (convex lens)
m = +3 (virtual, erect, magnified)

(a) Finding u:
From \( m = \frac{v}{u} \):
\[ +3 = \frac{v}{u} \implies v = 3u \] Since u is negative (let u = βˆ’x, where x > 0):
\[ v = 3 \times (-x) = -3x \] (v is negative β†’ virtual image on same side as object βœ“)

Using the lens formula:
\[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \] \[ \frac{1}{15} = \frac{1}{-3x} – \frac{1}{-x} = \frac{-1}{3x} + \frac{1}{x} = \frac{-1 + 3}{3x} = \frac{2}{3x} \] \[ 3x = 30 \implies x = 10 \] \[ u = -10 \text{ cm} \] Object distance = βˆ’10 cm (10 cm to the left, within the focal length βœ“).

(b) Image distance:
\[ v = 3u = 3 \times (-10) = -30 \text{ cm} \] Image distance = βˆ’30 cm (virtual, same side as object). βœ“

(c) Power:
\[ P = \frac{1}{f \text{ (in m)}} = \frac{1}{0.15} = +6.67 \text{ D} \] Power β‰ˆ +6.67 D

Question 4

Two thin lenses, a convex lens of power +5 D and a concave lens of power βˆ’3 D, are placed in contact. An object of height 4 cm is placed 60 cm from this combination. Find:
(a) the focal length of the combination,
(b) the position of the image,
(c) the magnification, and
(d) the height and nature of the image.

Hint: Net power P = P₁ + Pβ‚‚. Compute combined f from P. Then apply sign convention: u = βˆ’60 cm. Combined f will be positive (since net power is positive β†’ acts as convex). Use the lens formula to find v.

Show Solution
Given:
P₁ = +5 D (convex), Pβ‚‚ = βˆ’3 D (concave)
h = +4 cm, u = βˆ’60 cm

(a) Combined focal length:
\[ P = P_1 + P_2 = +5 + (-3) = +2 \text{ D} \] \[ f = \frac{1}{P} = \frac{1}{2} = 0.50 \text{ m} = +50 \text{ cm} \] Combined focal length = +50 cm (acts as a convex lens). βœ“

(b) Image position:
\[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \] \[ \frac{1}{50} = \frac{1}{v} – \frac{1}{-60} = \frac{1}{v} + \frac{1}{60} \] \[ \frac{1}{v} = \frac{1}{50} – \frac{1}{60} = \frac{6 – 5}{300} = \frac{1}{300} \] \[ v = +300 \text{ cm} \] Image is formed at +300 cm (3 m to the right of the lens β†’ real image). βœ“

(c) Magnification:
\[ m = \frac{v}{u} = \frac{300}{-60} = -5 \] Magnification = βˆ’5 (negative β†’ inverted; |m| > 1 β†’ magnified).

(d) Image height:
\[ m = \frac{h’}{h} \implies h’ = m \times h = (-5) \times 4 = -20 \text{ cm} \] Image height = βˆ’20 cm (negative β†’ below principal axis β†’ inverted).

Nature: The image is real, inverted, and magnified.

Question 5

An object is placed at a distance of 40 cm from a concave lens of focal length 20 cm. An erect image of height 2 cm is obtained. Find:
(a) the image distance,
(b) the magnification,
(c) the height of the object, and
(d) the power of the lens.

Hint: Concave lens β†’ f = βˆ’20 cm, u = βˆ’40 cm. Concave lens always forms virtual, erect images β†’ v will be negative, hβ€² will be positive. Do NOT accidentally put f as positive!

Show Solution
Given:
f = βˆ’20 cm, u = βˆ’40 cm, hβ€² = +2 cm (erect)

(a) Image distance:
\[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \] \[ \frac{1}{-20} = \frac{1}{v} – \frac{1}{-40} = \frac{1}{v} + \frac{1}{40} \] \[ \frac{1}{v} = \frac{1}{-20} – \frac{1}{40} = \frac{-2 – 1}{40} = \frac{-3}{40} \] \[ v = \frac{-40}{3} = -13.33 \text{ cm} \] Image distance = βˆ’13.33 cm (negative β†’ virtual, on same side as object). οΏ½οΏ½οΏ½

(b) Magnification:
\[ m = \frac{v}{u} = \frac{-40/3}{-40} = +\frac{1}{3} \approx +0.33 \] Magnification = +1/3 (positive β†’ erect; |m| < 1 β†’ diminished). βœ“

(c) Height of the object:
\[ m = \frac{h’}{h} \implies h = \frac{h’}{m} = \frac{2}{1/3} = 6 \text{ cm} \] Object height = 6 cm

(d) Power:
\[ P = \frac{1}{f \text{ (in m)}} = \frac{1}{-0.20} = -5 \text{ D} \] Power = βˆ’5 D (negative confirms diverging/concave lens). βœ“

Question 6

A convex lens forms a real image at a distance of 60 cm from the lens. The image is 2 times the size of the object and is inverted. Find:
(a) the object distance,
(b) the focal length,
(c) the power of the lens, and
(d) if the object is now moved 10 cm closer to the lens, find the new image distance.

Hint: Real & inverted image β†’ m = βˆ’2, v = +60 cm. Find u from m = v/u. For part (d), the new u changes β€” recalculate v using the same f. Be careful: moving the object closer means |u| decreases (u becomes less negative).

Show Solution
Given:
v = +60 cm (real image), m = βˆ’2 (inverted, magnified)

(a) Object distance:
\[ m = \frac{v}{u} \implies -2 = \frac{60}{u} \implies u = \frac{60}{-2} = -30 \text{ cm} \] Object distance = βˆ’30 cm. βœ“

(b) Focal length:
\[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{60} – \frac{1}{-30} = \frac{1}{60} + \frac{1}{30} = \frac{1 + 2}{60} = \frac{3}{60} = \frac{1}{20} \] \[ f = +20 \text{ cm} \] Focal length = +20 cm. βœ“

(c) Power:
\[ P = \frac{1}{0.20} = +5 \text{ D} \]
(d) Object moved 10 cm closer:
Original |u| = 30 cm. Moving 10 cm closer β†’ new |u| = 20 cm.
New u = βˆ’20 cm, f = +20 cm.
\[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \] \[ \frac{1}{20} = \frac{1}{v} – \frac{1}{-20} = \frac{1}{v} + \frac{1}{20} \] \[ \frac{1}{v} = \frac{1}{20} – \frac{1}{20} = 0 \] \[ v = \infty \] When the object is at the focus (u = βˆ’f), the image is formed at infinity.
The rays emerge parallel, and no finite image is formed. This is the principle behind a searchlight / collimator.

Question 7

A student claims that a concave lens of focal length 15 cm formed a real, inverted image at +30 cm when an object was placed at 30 cm from the lens. Verify this claim using the lens formula. What image will actually be formed?

Hint: This is a real vs virtual image trap. Concave lens β†’ f = βˆ’15 cm. Object distance β†’ u = βˆ’30 cm. Apply the lens formula and check whether v comes out positive or negative. A concave lens CANNOT form a real image of a real object.

Show Solution
Given (as per student’s claim):
f = βˆ’15 cm (concave lens), u = βˆ’30 cm

Verification using Lens Formula:
\[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \] \[ \frac{1}{-15} = \frac{1}{v} – \frac{1}{-30} = \frac{1}{v} + \frac{1}{30} \] \[ \frac{1}{v} = \frac{1}{-15} – \frac{1}{30} = \frac{-2 – 1}{30} = \frac{-3}{30} = \frac{-1}{10} \] \[ v = -10 \text{ cm} \] Actual result:
v = βˆ’10 cm (not +30 cm as claimed).

The student’s claim is WRONG. βœ—

The image is formed at 10 cm on the same side as the object.

Magnification:
\[ m = \frac{v}{u} = \frac{-10}{-30} = +\frac{1}{3} \] The image is virtual, erect, and diminished (to 1/3 of object size).

Key Concept: A concave lens always forms a virtual, erect, and diminished image for a real object, regardless of the object’s position. It can never form a real image of a real object.

Question 8

A convex lens of focal length 12 cm produces a real image that is located 4 times as far from the lens as the object. Find:
(a) the object distance,
(b) the image distance,
(c) the magnification, and
(d) if the object height is 3 cm, find the image height and state whether it is erect or inverted.

Hint: “Image is 4 times as far” means |v| = 4|u|. Since it’s real β†’ v = +4|u|. Object is on the left β†’ u = βˆ’|u|. The magnification’s sign will reveal if it’s erect or inverted. Careful with the ratio β€” it gives |v|/|u|, NOT magnification directly.

Show Solution
Given:
f = +12 cm (convex lens)
Real image, so v is positive.
|v| = 4|u|

Let |u| = x, so u = βˆ’x and v = +4x.

(a) & (b) Object and Image distance:
Using the lens formula:
\[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{4x} – \frac{1}{-x} = \frac{1}{4x} + \frac{1}{x} = \frac{1 + 4}{4x} = \frac{5}{4x} \] \[ \frac{1}{12} = \frac{5}{4x} \] \[ 4x = 60 \implies x = 15 \] \[ u = -15 \text{ cm}, \quad v = +60 \text{ cm} \] Object distance = βˆ’15 cm, Image distance = +60 cm. βœ“

(c) Magnification:
\[ m = \frac{v}{u} = \frac{60}{-15} = -4 \] Magnification = βˆ’4 (negative β†’ inverted; |m| = 4 β†’ magnified). βœ“

(d) Image height:
\[ h’ = m \times h = (-4) \times 3 = -12 \text{ cm} \] Image height = βˆ’12 cm.
Negative sign β†’ the image is inverted (below the principal axis).
The image is real, inverted, and magnified.

Question 9

A lens has a power of βˆ’4 D. An object of height 6 cm is placed at a distance of 50 cm from the lens. Find:
(a) the type and focal length of the lens,
(b) the image distance,
(c) the magnification, and
(d) the height and nature of the image.

Also explain: Can this lens ever produce a magnified image?

Hint: Negative power β†’ concave lens β†’ f is negative. u = βˆ’50 cm. Expect v to be negative (virtual image). For the “can it magnify?” part, think about what a diverging lens always does to light rays from real objects.

Show Solution
Given:
P = βˆ’4 D, h = +6 cm, u = βˆ’50 cm

(a) Type and focal length:
\[ f = \frac{1}{P} = \frac{1}{-4} = -0.25 \text{ m} = -25 \text{ cm} \] Negative focal length β†’ Concave (diverging) lens.
Focal length = βˆ’25 cm. βœ“

(b) Image distance:
\[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \] \[ \frac{1}{-25} = \frac{1}{v} – \frac{1}{-50} = \frac{1}{v} + \frac{1}{50} \] \[ \frac{1}{v} = \frac{1}{-25} – \frac{1}{50} = \frac{-2 – 1}{50} = \frac{-3}{50} \] \[ v = \frac{-50}{3} = -16.67 \text{ cm} \] Image distance = βˆ’16.67 cm (virtual, same side as object). βœ“

(c) Magnification:
\[ m = \frac{v}{u} = \frac{-50/3}{-50} = +\frac{1}{3} \approx +0.33 \] Magnification = +1/3 (positive β†’ erect; diminished). βœ“

(d) Image height:
\[ h’ = m \times h = \frac{1}{3} \times 6 = +2 \text{ cm} \] Image height = +2 cm (positive β†’ erect, above principal axis).

Nature: The image is virtual, erect, and diminished.

Can a concave lens ever produce a magnified image?
No. A concave lens always diverges light rays. For any real object placed at any distance, it always produces a virtual, erect, and diminished image with |m| < 1. It can never produce a magnified image of a real object.

Question 10

An object of height 8 cm is placed at a distance of 25 cm from a convex lens of focal length 10 cm. A student calculates the image to be virtual and erect. Another student says the image is real and inverted.

(a) Using the lens formula, find the image distance and determine which student is correct.
(b) Find the magnification and height of the image.
(c) Calculate the power of the lens.
(d) What would happen to the image if the object were moved to a distance of 10 cm from the lens? Explain.

Hint: f = +10 cm, u = βˆ’25 cm. Since |u| > f, the object is beyond the focus β†’ expect a real image (v positive). For part (d), placing the object at F means |u| = f β€” recall what happens at the focus of a convex lens. This question tests conceptual reasoning + sign convention + special cases all in one.

Show Solution
Given:
h = +8 cm, u = βˆ’25 cm, f = +10 cm

(a) Image distance:
\[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \] \[ \frac{1}{10} = \frac{1}{v} – \frac{1}{-25} = \frac{1}{v} + \frac{1}{25} \] \[ \frac{1}{v} = \frac{1}{10} – \frac{1}{25} = \frac{5 – 2}{50} = \frac{3}{50} \] \[ v = +\frac{50}{3} = +16.67 \text{ cm} \] v is positive β†’ image is on the opposite side of the object β†’ real image.

βœ… The second student is correct. The image is real and inverted.
βœ— The first student is wrong.

Why? Since the object is placed at 25 cm, which is beyond F (f = 10 cm), a convex lens forms a real, inverted image. A convex lens produces a virtual image only when the object is between the optical centre and F (i.e., |u| < f).

(b) Magnification and image height:
\[ m = \frac{v}{u} = \frac{50/3}{-25} = \frac{50}{3 \times (-25)} = \frac{50}{-75} = -\frac{2}{3} \] Magnification = βˆ’2/3 (negative β†’ inverted; |m| < 1 β†’ diminished).

\[ h’ = m \times h = \left(-\frac{2}{3}\right) \times 8 = -\frac{16}{3} = -5.33 \text{ cm} \] Image height = βˆ’5.33 cm (negative β†’ inverted, below principal axis).

(c) Power:
\[ P = \frac{1}{f \text{ (in m)}} = \frac{1}{0.10} = +10 \text{ D} \] Power = +10 D. βœ“

(d) Object at 10 cm from the lens (u = βˆ’10 cm):
Since f = +10 cm, we have |u| = f, i.e., the object is at the focus.
\[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \] \[ \frac{1}{10} = \frac{1}{v} – \frac{1}{-10} = \frac{1}{v} + \frac{1}{10} \] \[ \frac{1}{v} = \frac{1}{10} – \frac{1}{10} = 0 \] \[ v = \infty \] When the object is placed at the principal focus of a convex lens, the refracted rays emerge parallel and the image is formed at infinity.

The image is:
β€’ highly magnified (theoretically infinitely large),
β€’ real,
β€’ formed at infinity.

This is the principle used in spotlights, searchlights, and collimated beam projectors.

πŸ“Š Set 4 – Summary Table

Q# Lens Type u (cm) f (cm) v (cm) m Image Nature Key Trap / Concept
1 Convex βˆ’30 +20 +60 βˆ’2 Real, Inverted, Magnified Multi-step: m β†’ v β†’ f β†’ P
2 Concave βˆ’60 βˆ’20 βˆ’15 +1/4 Virtual, Erect, Diminished Finding u from v and f
3 Convex βˆ’10 +15 βˆ’30 +3 Virtual, Erect, Magnified Virtual image by convex lens (magnifying glass)
4 Combination βˆ’60 +50 +300 βˆ’5 Real, Inverted, Magnified Power combination + multi-step
5 Concave βˆ’40 βˆ’20 βˆ’13.33 +1/3 Virtual, Erect, Diminished Finding object height from image height
6 Convex βˆ’30 +20 +60 β†’ ∞ βˆ’2 β†’ ∞ Real β†’ Image at infinity Object moved to focus (v = ∞)
7 Concave βˆ’30 βˆ’15 βˆ’10 +1/3 Virtual, Erect, Diminished Real vs virtual image trap
8 Convex βˆ’15 +12 +60 βˆ’4 Real, Inverted, Magnified Distance ratio β‰  magnification trap
9 Concave βˆ’50 βˆ’25 βˆ’16.67 +1/3 Virtual, Erect, Diminished Power β†’ f β†’ lens type + conceptual
10 Convex βˆ’25 +10 +16.67 βˆ’2/3 Real, Inverted, Diminished Student debate + special case at F

πŸ“˜ Class 10 Physics – Lenses Numerical Practice | Set 4 – Board-Level Tricky Problems

Prepared for CBSE Board Exam Practice | New Cartesian Sign Convention

Next: Set 5 – Mixed Problems with HOTS & Assertion-Reason

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