Numericals On Lens
For convex lens
LEVEL-1 to LEVEL-2 (Q1–Q7): Conceptual + Calculation Accuracy U = (-)
Q1) An object is placed 30 cm in front of a convex lens of focal length 15 cm.
1. Find the image distance.
2. Determine the nature and magnification of the image.
Q2) A concave lens of focal length −20 cm forms an image 10 cm from the lens.
1. Find the object distance.
2. Calculate the magnification.
Q3) An object is placed at twice the focal length of a convex lens.
1. Where is the image formed?
2. What is the magnification?
3. If the object is moved 5 cm towards the lens, what happens to the image?
Q4) A convex lens produces a real image three times smaller than the object at a distance of 60 cm from
the lens.
1. Find the focal length of the lens.
2. Find the object distance.
Q5) An object is placed between F and 2F of a convex lens.
1. Prove mathematically that the magnification is greater than 1.
2. Show that the image is always formed beyond 2F.
Q6) Two convex lenses of focal lengths 20 cm and 30 cm are placed in contact.
1. Find the equivalent focal length.
2. If an object is placed 25 cm from the combination, find the final image position.
Q7) A convex lens forms an image on a screen. When the lens is shifted 5 cm away from the screen, the
image becomes blurred.
1. In which direction should the screen be moved to regain focus?
2. Explain using lens formula (no ray diagram).
EXTREMELY TOUGH & TRICKIEST PROBLEMS (Q8–Q15)
Q8 (Concept Trap)
An object is placed at a distance of u from a convex lens. The magnification is m.
1. Derive an expression for focal length f in terms of u and m.
2. For what value of m does the image go to infinity?
Q9 (Hidden Sign Convention Trap)
A lens produces a real image of an object placed at 20 cm. When the object is moved 5 cm closer, the
image becomes virtual and erect.
1. Find the focal length of the lens.
2. Explain why a small shift causes a complete change in image nature.
Q10 (Two-Lens Extreme Trap)
Two thin lenses L₁ (f₁ = +10 cm) and L₂ (f₂ = −20 cm) are separated by 15 cm.
An object is placed 5 cm to the left of L₁.
1. Find the final image position.
2. Determine the net magnification.
3. State whether the final image can be caught on a screen.
, Q11 (Image Coincidence – Killer Problem)
An object and a screen are fixed 90 cm apart. A convex lens is placed between them to form a sharp image
on the screen.
1. Find the two possible positions of the lens.
2. If the focal length of the lens is 20 cm, calculate the distance between the two lens positions.
Q12 (Relative Motion of Object & Lens)
An object moves towards a convex lens with a speed v₀. The lens moves away from the object with speed
v₀/2.
1. Determine whether the image approaches or recedes from the lens.
2. Under what condition does the image remain stationary?
Q13 (Magnification Extremum – Very Hard)
For a convex lens, show that the magnification is minimum when the object is placed at 2f.
1. Prove mathematically.
2. Find the minimum magnification.
Q14 (Lens Removed Mid-System – Olympiad Level)
Two convex lenses of focal lengths 10 cm and 20 cm are placed 30 cm apart.
An object is placed 15 cm from the first lens.
1. Find the final image position.
2. If the second lens is suddenly removed, where does the image shift?
3. By how much must the screen be moved?
Q15 (EXTREME — Multiple Constraints)
An object is placed in front of a convex lens and produces an image of same size as the object.
Now the object is moved 10 cm closer, and the lens is moved so that:
the image remains real,
the image remains at the same position,
magnification becomes 2.
1. Find the focal length of the lens.
2. Find the distance through which the lens is shifted.
3. Explain physically why all three conditions can be satisfied simultaneously.
ANSWERS: LENS NUMERICALS (Q1–Q15)
Q1) Image distance: +30 cm Q6) Equivalent focal length: 12 cm Q11) Two lens positions: 30 cm and 60 cm from
Nature: Real, inverted Final image position: 23.08 cm object
Magnification: −1 (virtual) Distance between positions: 30 cm
Q2) Object distance: −20 cm Q7) Screen should be moved towards the Q12) Image recedes from lens
Magnification: +0.5 lens Image remains stationary when:
Image distance decreases du dv
=
Q3) Image position: 2f
EXTREMELY TOUGH (Q8–Q15)
dt dt
Magnification: −1
After shifting object 5 cm towards Q13) Minimum magnification at: u = −2f
lens: Image moves beyond 2f, um Minimum magnification: 1
magnification increases Q8) f=
m−1
Image at infinity when m = 1 Q14) Final image position: 10 cm right of second lens
Q4) Focal length: 45 cm New image position (after removing L₂): 20
Object distance: −180 cm Q9) Focal length: 10 cm cm right of L₁
Reason: Object crosses focal point Screen shift: 10 cm
Q5) Magnification: > 1
Image position: Beyond 2F Q10) Final image position: 10 cm left of L₂ Q15) Focal length: 20 cm
Net magnification: −1 Lens shift: 10 cm
IMPOSSIBLE-LOOKING LENS PROBLEMS (Q1–Q10) All conditions satisfied at u = −40 cm
initially
For convex lens
LEVEL-1 to LEVEL-2 (Q1–Q7): Conceptual + Calculation Accuracy U = (-)
Q1) An object is placed 30 cm in front of a convex lens of focal length 15 cm.
1. Find the image distance.
2. Determine the nature and magnification of the image.
Q2) A concave lens of focal length −20 cm forms an image 10 cm from the lens.
1. Find the object distance.
2. Calculate the magnification.
Q3) An object is placed at twice the focal length of a convex lens.
1. Where is the image formed?
2. What is the magnification?
3. If the object is moved 5 cm towards the lens, what happens to the image?
Q4) A convex lens produces a real image three times smaller than the object at a distance of 60 cm from
the lens.
1. Find the focal length of the lens.
2. Find the object distance.
Q5) An object is placed between F and 2F of a convex lens.
1. Prove mathematically that the magnification is greater than 1.
2. Show that the image is always formed beyond 2F.
Q6) Two convex lenses of focal lengths 20 cm and 30 cm are placed in contact.
1. Find the equivalent focal length.
2. If an object is placed 25 cm from the combination, find the final image position.
Q7) A convex lens forms an image on a screen. When the lens is shifted 5 cm away from the screen, the
image becomes blurred.
1. In which direction should the screen be moved to regain focus?
2. Explain using lens formula (no ray diagram).
EXTREMELY TOUGH & TRICKIEST PROBLEMS (Q8–Q15)
Q8 (Concept Trap)
An object is placed at a distance of u from a convex lens. The magnification is m.
1. Derive an expression for focal length f in terms of u and m.
2. For what value of m does the image go to infinity?
Q9 (Hidden Sign Convention Trap)
A lens produces a real image of an object placed at 20 cm. When the object is moved 5 cm closer, the
image becomes virtual and erect.
1. Find the focal length of the lens.
2. Explain why a small shift causes a complete change in image nature.
Q10 (Two-Lens Extreme Trap)
Two thin lenses L₁ (f₁ = +10 cm) and L₂ (f₂ = −20 cm) are separated by 15 cm.
An object is placed 5 cm to the left of L₁.
1. Find the final image position.
2. Determine the net magnification.
3. State whether the final image can be caught on a screen.
, Q11 (Image Coincidence – Killer Problem)
An object and a screen are fixed 90 cm apart. A convex lens is placed between them to form a sharp image
on the screen.
1. Find the two possible positions of the lens.
2. If the focal length of the lens is 20 cm, calculate the distance between the two lens positions.
Q12 (Relative Motion of Object & Lens)
An object moves towards a convex lens with a speed v₀. The lens moves away from the object with speed
v₀/2.
1. Determine whether the image approaches or recedes from the lens.
2. Under what condition does the image remain stationary?
Q13 (Magnification Extremum – Very Hard)
For a convex lens, show that the magnification is minimum when the object is placed at 2f.
1. Prove mathematically.
2. Find the minimum magnification.
Q14 (Lens Removed Mid-System – Olympiad Level)
Two convex lenses of focal lengths 10 cm and 20 cm are placed 30 cm apart.
An object is placed 15 cm from the first lens.
1. Find the final image position.
2. If the second lens is suddenly removed, where does the image shift?
3. By how much must the screen be moved?
Q15 (EXTREME — Multiple Constraints)
An object is placed in front of a convex lens and produces an image of same size as the object.
Now the object is moved 10 cm closer, and the lens is moved so that:
the image remains real,
the image remains at the same position,
magnification becomes 2.
1. Find the focal length of the lens.
2. Find the distance through which the lens is shifted.
3. Explain physically why all three conditions can be satisfied simultaneously.
ANSWERS: LENS NUMERICALS (Q1–Q15)
Q1) Image distance: +30 cm Q6) Equivalent focal length: 12 cm Q11) Two lens positions: 30 cm and 60 cm from
Nature: Real, inverted Final image position: 23.08 cm object
Magnification: −1 (virtual) Distance between positions: 30 cm
Q2) Object distance: −20 cm Q7) Screen should be moved towards the Q12) Image recedes from lens
Magnification: +0.5 lens Image remains stationary when:
Image distance decreases du dv
=
Q3) Image position: 2f
EXTREMELY TOUGH (Q8–Q15)
dt dt
Magnification: −1
After shifting object 5 cm towards Q13) Minimum magnification at: u = −2f
lens: Image moves beyond 2f, um Minimum magnification: 1
magnification increases Q8) f=
m−1
Image at infinity when m = 1 Q14) Final image position: 10 cm right of second lens
Q4) Focal length: 45 cm New image position (after removing L₂): 20
Object distance: −180 cm Q9) Focal length: 10 cm cm right of L₁
Reason: Object crosses focal point Screen shift: 10 cm
Q5) Magnification: > 1
Image position: Beyond 2F Q10) Final image position: 10 cm left of L₂ Q15) Focal length: 20 cm
Net magnification: −1 Lens shift: 10 cm
IMPOSSIBLE-LOOKING LENS PROBLEMS (Q1–Q10) All conditions satisfied at u = −40 cm
initially
