Ray Optics · Complete Exam Notes

Lens Formula, Derivation, Magnification and Image Formation

A visual, sign-correct chapter for CBSE, NEET, IIT-JEE, AP Physics, IB, IGCSE and A-Level students.

1. Lens Formula

For a thin lens under paraxial approximation, using the Cartesian sign convention:

1/f = 1/v − 1/u

u: signed object distance from optical centre O.

v: signed image distance from O.

f: signed focal length; positive for convex and negative for concave lenses.

Exam Tip: Substitute signed values. Do not insert an extra negative sign for u after already writing u as negative.

2. Three Case-wise Derivations

Case 1: Convex Lens — Real, Inverted Image

Question: For a convex lens with object AB between F₁ and 2F₁, derive the lens formula and linear magnification using similar triangles.

Given Data: u=−U, v=+V, f>0, h>0 and h′<0.

Formula Used: Similar-triangle ratios and m=h′/h.

Step-by-Step Solution:

From similar triangles ABO and A′B′O: h′/h=v/u.
For the ray parallel to the axis, the refracted segment and the focal triangle give: (h′−h)/v=−h/f.
Put h′=hv/u: (hv/u−h)/v=−h/f.
Therefore 1/u−1/v=−1/f.
Hence 1/f=1/v−1/u.
m=h′/h=v/u<0, so the image is inverted.

Final Answer: 1/f = 1/v − 1/u | m = v/u

Object between F₁ and 2F₁; image beyond 2F₂

Common Mistake: Drawing a real image upright or placing it on the object side.

Exam Tip: Verify that the parallel incident ray passes through F₂ after refraction.

Case 2: Convex Lens — Virtual, Erect Image

Question: For an object AB between O and F₁ of a convex lens, derive the lens formula and magnification and explain the positive sign of m.

Given Data: u<0, v<0, f>0 and h′>0.

Formula Used: Directed similar triangles and m=h′/h.

Step-by-Step Solution:

Directed similar triangles through O give h′/h=v/u.
The focal ray geometry still gives (h′−h)/v=−h/f.
Substitute h′=hv/u and divide by h.
1/u−1/v=−1/f.
Thus 1/f=1/v−1/u.
Since v and u are both negative, m=v/u>0; the image is erect and magnified.

Final Answer: f>0, u<0, v<0, m>1

Object inside focal length; virtual magnified image

Common Mistake: Extending the real green rays forward to invent a real image.

Exam Tip: The central ray is undeviated; both dotted extensions must meet exactly at B′.

Case 3: Concave Lens — Virtual, Erect, Diminished Image

Question: For a real object in front of a concave lens, derive the lens formula and prove that magnification is positive but less than one.

Given Data: f<0, u<0, v<0 and |v|<|u|.

Formula Used: Directed similar triangles and m=v/u.

Step-by-Step Solution:

Central-ray triangles give h′/h=v/u.
Using the directed focal triangle: (h′−h)/v=−h/f.
Substitution again gives 1/u−1/v=−1/f.
Therefore 1/f=1/v−1/u.
Both v and u are negative, so m>0.
Because |v|<|u|, 0<m<1: virtual, erect and diminished, never negative.

Final Answer: f<0, v<0, 0<m<1

Concave lens virtual-image construction

Common Mistake: Giving a concave lens negative magnification for a real object.

Exam Tip: The parallel ray must appear to originate from F₁; the optical-centre ray remains straight.

3. Cartesian Sign Convention

Quantity / Situation	Sign	Reason
Origin	Optical centre O	All axial distances are measured from O.
Direction of incident light	Positive	Usually left to right.
Opposite direction	Negative	Left of O for left-to-right light.
Real object u	Negative	Object lies left of O.
Convex-lens real image v	Positive	Image lies right of O.
Convex/concave virtual image v	Negative	Image lies on object side.
Convex lens f	Positive	Principal focus is real.
Concave lens f	Negative	Principal focus is virtual.
Real inverted image m	Negative	h′ has opposite sign to h.
Virtual erect image m	Positive	h′ and h have the same sign.

Common Mistake: “Real means positive” is not a universal sign rule. The sign follows position and direction, not the word real alone.

4. Magnification and Linear Magnification

m = h′/h = v/u

m>0: erect image.

m<0: inverted image.

|m|>1: enlarged image.

|m|<1: diminished image.

|m|=1: same-size image.

Linear magnification: ratio of image height to object height; it has no unit.

5. Power of a Lens and Focal Length

P = 1/f(m)

Power measures the convergence or divergence produced by a lens. Its SI unit is the dioptre (D).

Convex lens: positive power.
Concave lens: negative power.
More curvature generally means more power.
Larger |f| means smaller |P|.

P(D) = 100/f(cm)
f(cm) = 100/P(D)

Fast check: A +5 D lens has f=+20 cm; a −2 D lens has f=−50 cm.

6. Image Formation Cases

Convex lens is converging in air, but it may behave as a diverging lens only when the surrounding medium has refractive index greater than the lens material.

Convex Lens

Object position	Image position	Nature	Size	m
Infinity	F₂	Real, inverted	Highly diminished	Negative, near 0
Beyond 2F₁	F₂ to 2F₂	Real, inverted	Diminished	−, \|m\|<1
At 2F₁	At 2F₂	Real, inverted	Same	−1
F₁ to 2F₁	Beyond 2F₂	Real, inverted	Enlarged	−, \|m\|>1
At F₁	Infinity	Real limit	Highly enlarged	Very large negative
O to F₁	Object side	Virtual, erect	Enlarged	Positive >1

Object at Infinity

Parallel rays meet at F₂.

Object Beyond 2F₁

Image lies between F₂ and 2F₂.

Object at 2F₁

Same-size image at 2F₂.

Object Between F₁ and 2F₁

Enlarged image beyond 2F₂.

Object at F₁

Emergent rays are parallel; image at infinity.

Object Between O and F₁

Virtual, erect, enlarged image on object side.

Concave Lens

Concave lens is diverging in air and forms a virtual, erect and diminished image for a real object.

Object at infinity: image at F₁; virtual, erect and highly diminished; m positive and near zero.

Object at finite distance: image between O and F₁; virtual, erect and diminished; 0<m<1.

Concave: Object at Infinity

Concave: Finite Object

7. Lens in Water or Another Medium

1/f_medium = (μ_lens/μ_medium − 1)(1/R₁ − 1/R₂)

As surrounding index increases toward the lens index, optical power decreases and |f| increases.

If μ_medium=μ_lens, power is zero and f→∞.

If μ_medium>μ_lens, a convex lens behaves as a diverging lens.

Mirror in Water: Important Comparison

Lens: focal length changes in water because refraction depends on relative refractive index.

Mirror: focal length does not change in water because reflection and mirror geometry do not depend on surrounding refractive index.

8. Twenty Fully Solved Numerical Problems

1. Convex lens: f=20 cm, u=−30 cm

Question: A convex lens of focal length 20 cm has an object placed 30 cm in front of it. Find the image distance and magnification.

Given: f=+20 cm, u=−30 cm.

Formula: 1/f=1/v−1/u.

Substitution: 1/20=1/v+1/30.

Calculation: 1/v=1/60; v=+60 cm; m=60/(−30)=−2.

Image at 60 cm; real, inverted, 2× enlarged.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

2. Convex lens: f=15 cm, u=−45 cm

Question: An object is placed 45 cm in front of a convex lens of focal length 15 cm. Find the image position and magnification.

Given: f=+15 cm, u=−45 cm.

Formula: 1/v=1/f+1/u.

Substitution: 1/v=1/15−1/45.

Calculation: 1/v=2/45; v=22.5 cm; m=−0.5.

Real, inverted image at 22.5 cm; half-size.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

3. Object at 2F of a 10 cm convex lens

Question: An object is placed at 2F₁ of a convex lens whose focal length is 10 cm. Find the image position, nature and size.

Given: f=+10 cm, u=−20 cm.

Formula: 1/f=1/v−1/u.

Substitution: 1/10=1/v+1/20.

Calculation: v=+20 cm; m=−1.

Image at 2F₂, real, inverted and same size.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

4. Object inside focal length

Question: An object is placed 8 cm from a convex lens of focal length 12 cm. Find the image distance and magnification.

Given: f=+12 cm, u=−8 cm.

Formula: 1/v=1/f+1/u.

Substitution: 1/v=1/12−1/8.

Calculation: 1/v=−1/24; v=−24 cm; m=3.

Virtual erect image 24 cm away; 3× enlarged.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

5. Find object position

Question: A convex lens of focal length 10 cm forms a real image 30 cm from the lens. Find the object position.

Given: f=+10 cm, v=+30 cm.

Formula: 1/u=1/v−1/f.

Substitution: 1/u=1/30−1/10.

Calculation: 1/u=−1/15.

u=−15 cm.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

6. Concave lens: f=−15 cm, u=−30 cm

Question: An object is placed 30 cm in front of a concave lens of focal length 15 cm. Find the image position and magnification.

Given: f=−15 cm, u=−30 cm.

Formula: 1/v=1/f+1/u.

Substitution: 1/v=−1/15−1/30.

Calculation: v=−10 cm; m=1/3.

Virtual, erect, diminished image at 10 cm.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

7. Concave lens: f=−20 cm, u=−60 cm

Question: An object is placed 60 cm in front of a concave lens of focal length 20 cm. Find the image position and magnification.

Given: f=−20 cm, u=−60 cm.

Formula: 1/v=1/f+1/u.

Substitution: −1/20−1/60=−1/15.

Calculation: v=−15 cm; m=+0.25.

Virtual erect image; one-quarter size.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

8. Power of a 25 cm convex lens

Question: Calculate the power of a convex lens whose focal length is 25 cm.

Given: f=+25 cm=+0.25 m.

Formula: P=1/f(m).

Substitution: P=1/0.25.

Calculation: P=4 D.

Power = +4 D.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

9. Focal length from power −2.5 D

Question: A lens has power −2.5 D. Find its focal length and identify the lens type.

Given: P=−2.5 D.

Formula: f=1/P.

Substitution: f=1/(−2.5) m.

Calculation: f=−0.40 m=−40 cm.

Concave lens; f=−40 cm.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

10. Focal length of a +5 D lens

Question: A lens has power +5 D. Calculate its focal length and identify the lens type.

Given: P=+5 D.

Formula: f(cm)=100/P.

Substitution: f=100/5.

Calculation: f=20 cm.

Convex lens; f=+20 cm.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

11. Two lenses of +2 D and +3 D

Question: Two thin lenses of powers +2 D and +3 D are placed in contact. Find their equivalent power and focal length.

Given: P₁=+2 D, P₂=+3 D.

Formula: P=P₁+P₂; f=1/P.

Substitution: P=5 D.

Calculation: f=0.20 m.

Equivalent power +5 D; focal length +20 cm.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

12. Convex 20 cm and concave −50 cm in contact

Question: A convex lens of focal length 20 cm and a concave lens of focal length 50 cm are in contact. Find the equivalent focal length.

Given: f₁=+0.20 m, f₂=−0.50 m.

Formula: P=1/f₁+1/f₂.

Substitution: P=5−2.

Calculation: P=3 D; f=1/3 m.

Equivalent focal length = +33.3 cm.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

13. Magnification from u and v

Question: An image is formed 40 cm from a lens for an object 20 cm in front of it. Calculate linear magnification and state image orientation.

Given: u=−20 cm, v=+40 cm.

Formula: m=v/u.

Substitution: m=40/(−20).

Calculation: m=−2.

Image is inverted and twice the object height.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

14. Virtual magnification +2.5 at u=−12 cm

Question: A convex lens forms a virtual image with magnification +2.5 for an object 12 cm in front of it. Find v and f.

Given: m=+2.5, u=−12 cm.

Formula: v=mu; 1/f=1/v−1/u.

Substitution: v=2.5(−12)=−30 cm.

Calculation: 1/f=−1/30+1/12=1/20.

v=−30 cm and f=+20 cm.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

15. Real image with m=−0.5 at v=30 cm

Question: A real image has magnification −0.5 and is formed 30 cm from a convex lens. Find object distance and focal length.

Given: m=−0.5, v=+30 cm.

Formula: u=v/m; 1/f=1/v−1/u.

Substitution: u=30/(−0.5)=−60 cm.

Calculation: 1/f=1/30+1/60=1/20.

u=−60 cm and f=+20 cm.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

16. Glass lens in water

Question: A glass lens of refractive index 1.5 has focal length 20 cm in air. Find its focal length in water of refractive index 4/3.

Given: f_air=20 cm, μ_L=1.5, μ_w=4/3.

Formula: f_w/f_air=(μ_L−1)/(μ_L/μ_w−1).

Substitution: ratio=0.5/(1.125−1)=4.

Calculation: f_w=4×20.

Focal length in water = +80 cm.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

17. Lens and liquid have equal index

Question: What happens to the focal length of a lens when the refractive index of the surrounding liquid equals that of the lens?

Given: μ_L=μ_m.

Formula: 1/f=(μ_L/μ_m−1)K.

Substitution: relative-index factor=1−1=0.

Calculation: Power=0.

f=∞; the lens produces no deviation.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

18. Convex lens in a higher-index liquid

Question: A convex glass lens of refractive index 1.5 has focal length 20 cm in air. Find its focal length in a liquid of refractive index 1.6.

Given: f_air=20 cm, μ_L=1.5, μ_m=1.6.

Formula: f_m=f_air(μ_L−1)/(μ_L/μ_m−1).

Substitution: 20(0.5)/(0.9375−1).

Calculation: 10/(−0.0625)=−160 cm.

f=−160 cm; convex shape behaves as a concave lens.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

19. Distant object and 18 cm convex lens

Question: A distant object is viewed through a convex lens of focal length 18 cm. Where is the image formed?

Given: u=−∞, f=+18 cm.

Formula: 1/f=1/v−1/u.

Substitution: 1/18=1/v−0.

Calculation: v=+18 cm.

Image forms at F₂, 18 cm from the lens.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

20. Virtual image at 15 cm by a 10 cm convex lens

Question: A convex lens of focal length 10 cm forms a virtual image 15 cm from the lens. Find the object position and magnification.

Given: f=+10 cm, v=−15 cm.

Formula: 1/u=1/v−1/f.

Substitution: 1/u=−1/15−1/10=−1/6.

Calculation: u=−6 cm; m=(−15)/(−6)=2.5.

Object is 6 cm in front; image is virtual, erect and 2.5×.

Diagram: Use the matching convex-real, convex-virtual, concave-lens or power diagram given in the sections above.

Common Mistake: Substituting unsigned distances or mixing centimetres with metres.

Exam Tip: Predict the image type and sign before calculation.

9. Important PYQs and Exam-Oriented Questions

Exam-style questions based on previous exam patterns. Exact years are not claimed.

10 CBSE Subjective Questions

1. Derive the thin-lens formula for a convex lens forming a real image.

Answer: Similar triangles give m=h′/h=v/u and the focal triangle gives (h′−h)/v=−h/f. Eliminating heights yields 1/f=1/v−1/u. The real inverted image has m<0.

2. A +4 D lens forms an image with m=−2. Find u and v.

Answer: f=25 cm, v=−2u. Substitution gives u=−37.5 cm and v=+75 cm. The image is real and inverted.

3. State the Cartesian sign convention for lenses.

Answer: Measure from O; incident-light direction is positive; a real object has u<0; convex f>0; concave f<0; real-image v>0 and virtual-image v<0.

4. Why is the magnification of a concave lens positive?

Answer: Both u and v are negative, so m=v/u is positive. The image is erect; because |v|<|u|, m<1.

5. A convex lens of f=15 cm forms an image at 30 cm. Find u and m.

Answer: 1/15=1/30−1/u gives u=−30 cm and m=−1. Image is real, inverted and same size.

6. Explain image formation when an object is at F₁ of a convex lens.

Answer: The ray through O and the ray initially parallel to the axis emerge parallel. Therefore v→∞ and the image is at infinity.

7. Define one dioptre and find the power of a 50 cm concave lens.

Answer: One dioptre is the power of a lens with focal length 1 m. Here f=−0.50 m, so P=−2 D.

8. Compare a lens and a mirror when immersed in water.

Answer: Lens focal length changes because refraction depends on relative index. Mirror focal length remains unchanged because reflection depends on geometry.

9. Draw and describe a convex-lens virtual image.

Answer: Place the object between O and F₁. Refracted rays diverge; their dotted backward extensions meet on the object side, producing a virtual, erect, enlarged image.

10. Two lenses +5 D and −3 D are in contact. Find equivalent focal length.

Answer: P=+2 D, so f=1/2 m=+50 cm. The combination is converging.

15 NEET-Style MCQs

1. A concave lens can have magnification: A −2, B +2, C +0.4, D −0.4.

Correct answer: C. A real object gives a virtual erect diminished image, hence 0<m<1.

2. Powers +5 D and −2 D in contact give: A +7 D, B +3 D, C −3 D, D +2.5 D.

Correct answer: B. Powers add directly: P=5−2=3 D.

3. For a convex lens, v<0 means the object is: A beyond 2F, B at F, C between O and F, D at infinity.

Correct answer: C. Only an object inside the focal length produces a virtual image.

4. A lens of power −4 D has focal length: A +25 cm, B −25 cm, C −40 cm, D +40 cm.

Correct answer: B. f=100/P=−25 cm.

5. Object at 2F₁ of a convex lens gives m: A +1, B −1, C 0, D −2.

Correct answer: B. Image at 2F₂ is inverted and same size.

6. A ray through O of a thin lens: A bends toward axis, B bends away, C is undeviated, D reflects.

Correct answer: C. Refractions at the two surfaces cancel approximately for the optical-centre ray.

7. A convex lens has f=20 cm. Its power is: A +2 D, B +5 D, C −5 D, D +0.2 D.

Correct answer: B. P=100/20=+5 D.

8. A real image formed by a lens has magnification: A always positive, B always negative, C zero, D always greater than one.

Correct answer: B. A real image of a real object is inverted.

9. When μmedium=μlens, lens power is: A maximum, B zero, C negative, D unchanged.

Correct answer: B. Relative index is one, so the lens-maker factor is zero.

10. A parallel ray incident on a concave lens emerges: A through F₂, B parallel, C as if from F₁, D through 2F₂.

Correct answer: C. It diverges and its backward extension passes through F₁.

11. For u=−30 cm and v=+60 cm, m is: A +2, B −2, C +1/2, D −1/2.

Correct answer: B. m=v/u=60/(−30)=−2.

12. Object at F₁ of a convex lens forms image: A at F₂, B at 2F₂, C at infinity, D virtual at F₁.

Correct answer: C. Emergent rays are parallel.

13. Which always forms a virtual diminished image for a real object? A convex lens, B concave lens, C plane mirror, D concave mirror.

Correct answer: B. A concave lens always has v<0 and 0<m<1.

14. If |f| increases, |P|: A increases, B decreases, C is unchanged, D becomes zero always.

Correct answer: B. |P|=1/|f|.

15. A convex lens in a liquid denser optically than the lens behaves as: A converging, B diverging, C mirror, D no lens always.

Correct answer: B. μlens/μmedium−1 becomes negative.

10 JEE Main Questions

1. A lens has f=+20 cm and u=−30 cm. Find v and m.

Answer: v=+60 cm and m=−2.

2. A concave lens has f=−20 cm and u=−40 cm. Find v.

Answer: 1/v=−1/20−1/40=−3/40, so v=−13.33 cm.

3. A +10 D lens and −4 D lens are in contact. Find f.

Answer: P=+6 D, so f=1/6 m=16.67 cm.

4. A convex lens gives m=−3 for u=−20 cm. Find v and f.

Answer: v=+60 cm; 1/f=1/60+1/20=1/15, so f=15 cm.

5. A convex lens gives m=+4 for u=−9 cm. Find v and f.

Answer: v=−36 cm; 1/f=−1/36+1/9=1/12, so f=12 cm.

6. A 1.5-index lens has f=30 cm in air and is immersed in water (4/3). Find fwater.

Answer: Ratio=0.5/0.125=4, so fwater=120 cm.

7. Find power of a combination with focal lengths +25 cm and −100 cm.

Answer: P=+4−1=+3 D.

8. A lens forms a real same-size image 40 cm from it. Find f.

Answer: Same-size real image means u=−40 cm, v=+40 cm; f=20 cm.

9. An object at infinity gives image at x=+12 cm. Identify lens and power.

Answer: Convex lens, f=+12 cm, P=+8.33 D.

10. A lens in liquid has zero power. State the index condition.

Answer: μmedium=μlens, so relative refractive index equals one.

5 JEE Advanced Questions

1. A convex lens is moved through liquids of increasing index. Describe the sign changes of power.

Answer: Power decreases to zero at μmedium=μlens and becomes negative beyond that point. The physical convex lens then behaves as a diverging lens.

2. Two lenses f₁=+20 cm and f₂=−30 cm are separated by 10 cm. Find equivalent power.

Answer: P=P₁+P₂−dP₁P₂=5−3.333−0.1(5)(−3.333)=3.333 D; equivalent f≈30 cm.

3. A convex lens produces a virtual image twice the object size. Express u and v in terms of f.

Answer: m=2 gives v=2u. In the lens formula, 1/f=1/(2u)−1/u=−1/(2u), so u=−f/2 and v=−f.

4. A concave lens gives m=1/3 for u=−30 cm. Find f.

Answer: v=mu=−10 cm; 1/f=−1/10+1/30=−1/15, so f=−15 cm.

5. Why can a real object never give a real image through an isolated concave lens in air?

Answer: For f<0 and u<0, 1/v=1/f+1/u is always negative, hence v<0. Refracted rays diverge and only backward extensions meet.

5 IGCSE / A-Level / IB Questions

1. Describe an experiment to measure the focal length of a converging lens.

Answer: Form a sharp image of a distant object on a screen and measure lens-to-screen distance several times. Average the readings; distance approximates f.

2. A lens produces an upright image three times larger. Can it be projected?

Answer: No. Upright magnification is positive, indicating a virtual image, which cannot be caught on a screen.

3. Explain how to distinguish converging and diverging lenses using a distant object.

Answer: A converging lens forms a real focused image on a screen. A diverging lens cannot form such an image alone.

4. A +8 D lens is prescribed. Find focal length and lens type.

Answer: f=1/8 m=0.125 m=12.5 cm; positive power means converging lens.

5. Explain why the power of a glass lens decreases in water.

Answer: The relative index μglass/μwater is closer to one than μglass/μair, reducing refraction at each surface and therefore reducing power.

5 Assertion–Reason Questions

1. A: Concave-lens magnification is positive. R: Its image is virtual and erect.

Answer: Both true, and R correctly explains A.

2. A: A convex lens can behave as a diverging lens. R: This occurs if μmedium>μlens.

Answer: Both true, and R correctly explains A.

3. A: A mirror's focal length changes in water. R: Reflection depends on refractive index.

Answer: Both A and R are false. Mirror focal length depends on curvature.

4. A: Object at F₁ of a convex lens gives image at infinity. R: Emergent rays are parallel.

Answer: Both true, and R correctly explains A.

5. A: A real image has negative magnification. R: It is inverted relative to the object.

Answer: Both true, and R correctly explains A for a real object and a single thin lens.

3 Case-Study Questions

1. Camera focusing: f=5 cm; calculate v for u=−200 cm and u=−25 cm. Explain sensor motion.

Answer: v=5.13 cm for the distant object and 6.25 cm for the nearer object. The sensor must move farther from the lens for a nearer subject.

2. A glass lens (μ=1.5) works in air, water and a μ=1.6 liquid. Compare behavior.

Answer: It converges most strongly in air, weakly in water, and diverges in μ=1.6 liquid because the relative-index factor becomes negative.

3. A student uses a +5 D lens and a −2 D lens to build an optical system. Find net power and discuss image tendency.

Answer: In contact, net power is +3 D and f=33.3 cm. The system remains converging, though weaker than the +5 D lens alone.

Common Mistakes: mixing centimetres and metres in power, forgetting that u is negative for a real object, assigning negative magnification to a concave lens, and drawing virtual extensions as solid real rays.

Exam Strategy: First draw the axis and mark O, F₁, F₂ and 2F points. Predict the image quadrant and signs before using any formula.