Is it correct to say: Using the mirror formula 1/v + 1/u = 1/f for a lens?

For lenses: 1/v − 1/u = 1/f. Substitute u with its negative sign.

Is it correct to say: Calculating power with f in cm, e.g. P = 1/25?

P = 1/f with f in metres: f = 25 cm = 0.25 m ⇒ P = +4 D. Or use P = 100/f(cm).

Is it correct to say: Taking the focal length of a concave lens as positive?

Concave (diverging) lens: f and P are negative. Convex (converging) lens: f and P are positive.

Is it correct to say: Adding focal lengths for lenses in contact?

Add the POWERS (P = P₁ + P₂) or the reciprocals of focal lengths (1/f = 1/f₁ + 1/f₂).

Is it correct to say: Swapping the corrections for myopia and hypermetropia?

Myopia (near-sighted, image forms in front of the retina) ⇒ concave lens. Hypermetropia (far-sighted, image forms behind the retina) ⇒ convex lens.

Is it correct to say: Reporting a real image with positive magnification?

With Cartesian signs, a real image has v > 0 and u < 0, so M = v/u is negative (inverted).

📄 Formula sheet (PDF)

Lenses

Convex and concave lenses, Cartesian sign convention, lens formula, magnification, power, combination of lenses, the human eye and defects of vision — Maharashtra SSC Science-1 Ch 7

📐 10 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 10✓ 2025–26 syllabus

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Board Exam Tips

→Write the sign of every quantity before substituting: u is negative for a real object; f is positive for a convex lens and negative for a concave lens.
→After finding v, interpret it: v positive ⇒ real image on the other side of the lens; v negative ⇒ virtual image on the same side as the object.
→Power uses focal length in METRES. f = 25 cm ⇒ P = 1/0.25 = +4 D. Writing 1/25 is a common slip.
→Ray diagrams for a convex lens (object at infinity, beyond 2F₁, at 2F₁, between F₁ and 2F₁, at F₁, between F₁ and O) and for a concave lens must be neat, with arrows and labels.
→Learn the defects of vision as a table: defect, near/far point affected, cause, correcting lens.

📊 Diagram

Convex lens: a ray parallel to the principal axis passes through the principal focus F after refraction

📐 Formulas(10)

Cartesian Sign Convention for Lenses

u < 0 (real object), f_{convex} > 0, f_{concave} < 0

Lens Formula★ Board fav

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

Symbol	Meaning
$u$	Object distance from the optical centre (cm or m)
$v$	Image distance from the optical centre (cm or m)
$f$	Focal length (cm or m)

Magnification★ Board fav

M = \frac{h _{2}}{h _{1}} = \frac{v}{u}

Symbol	Meaning
$M$	Magnification (no unit)
$h_{1}$	Height of the object (cm)
$h_{2}$	Height of the image (cm)

Power of a Lens★ Board fav

P = \frac{1}{f ( in m )}

Symbol	Meaning
$P$	Power (dioptre, D)
$f$	Focal length in metres

Power with Focal Length in cm

P = \frac{100}{f ( in cm )}

Lenses in Contact: Focal Length

\frac{1}{f} = \frac{1}{f _{1}} + \frac{1}{f _{2}}

Symbol	Meaning
$f_{1}, f_{2}$	Focal lengths of the individual lenses
$f$	Equivalent focal length of the combination

Lenses in Contact: Power★ Board fav

P = P_{1} + P_{2} + P_{3} + \dots

Near Point and Far Point of a Normal Eye

near point = 25 cm, far point = \infty

Correction of Myopia (Near-sightedness)

f = - x (x = far point distance)

Symbol	Meaning
$x$	Distance of the defective eye's far point

Correction of Hypermetropia (Far-sightedness)

\frac{1}{f} = \frac{1}{- N} - \frac{1}{- 25} \Rightarrow f = \frac{25 N}{N - 25} (cm)

Symbol	Meaning
$N$	Near point of the defective eye (cm), N > 25 cm

✏️ Solved Examples

1Solved Exampleeasy2 steps

Find the power of a convex lens of focal length 40 cm.

Convert focal length to metres. Convex lens ⇒ f positive.

f = + 40 cm = + 0.4 m

2Solved Exampleboard4 steps

An object 4 cm high is placed 30 cm in front of a convex lens of focal length 20 cm. Find the position, size and nature of the image.

Cartesian signs: object on the left, convex lens.

u = - 30 cm, f = + 20 cm, h_{1} = + 4 cm

3Solved Exampleboard4 steps

An object is placed 30 cm from a concave lens of focal length 15 cm. Find the position and nature of the image.

Cartesian signs: concave lens ⇒ f negative.

u = - 30 cm, f = - 15 cm

4Solved ExampleHOTS4 steps

A convex lens of power +5 D and a concave lens of power −2 D are kept in contact. An object is placed 50 cm in front of the combination. Find the focal length of the combination and the position of the image.

Powers add for lenses in contact.

P = P_{1} + P_{2} = + 5 + (- 2) = + 3 D

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Using the mirror formula 1/v + 1/u = 1/f for a lens
✓For lenses: 1/v − 1/u = 1/f. Substitute u with its negative sign.
2
Calculating power with f in cm, e.g. P = 1/25
✓P = 1/f with f in metres: f = 25 cm = 0.25 m ⇒ P = +4 D. Or use P = 100/f(cm).
3
Taking the focal length of a concave lens as positive
✓Concave (diverging) lens: f and P are negative. Convex (converging) lens: f and P are positive.
4
Adding focal lengths for lenses in contact
✓Add the POWERS (P = P₁ + P₂) or the reciprocals of focal lengths (1/f = 1/f₁ + 1/f₂).
5
Swapping the corrections for myopia and hypermetropia
✓Myopia (near-sighted, image forms in front of the retina) ⇒ concave lens. Hypermetropia (far-sighted, image forms behind the retina) ⇒ convex lens.
6
Reporting a real image with positive magnification
✓With Cartesian signs, a real image has v > 0 and u < 0, so M = v/u is negative (inverted).

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
A doctor prescribes spectacles of power −2.5 D. Find the focal length. Which type of lens is it and which defect does it correct?
Q2
Three thin lenses of power +1.5 D, +2 D and −0.5 D are kept in contact. Find the total power and the focal length of the combination.
Q3
An object is placed 15 cm from a convex lens of focal length 10 cm. Find the image distance and magnification.
Q4
A convex lens of focal length 12 cm is used as a magnifying glass with the object 8 cm from it. Find the image position and magnification.
Q5
The far point of a myopic eye is 1.5 m. Find the power of the lens needed to correct it.
Q6
The near point of a hypermetropic eye is 75 cm. Find the focal length and power of the lens needed so that the person can read at 25 cm.

📝 Notes

Lenses

A lens is a transparent medium bounded by two surfaces, at least one of them curved. A convex lens is thicker at the middle and converges light; a concave lens is thinner at the middle and diverges it. This chapter turns that idea into three formulas and applies them to the human eye.

The Cartesian sign convention — decide signs first

Measure every distance from the optical centre O.
Light travels left to right: distances along it are positive, against it negative.
So a real object always has $u < 0$ ; a convex lens has $f > 0$ ; a concave lens has $f < 0$ .

With signs fixed, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ and $M = \frac{v}{u}$ tell you everything: the sign of $v$ gives the side of the image, the sign of $M$ tells whether it is inverted or erect.

Images formed by a convex lens

Object position	Image position	Nature
At infinity	At F₂	Real, inverted, point-sized
Beyond 2F₁	Between F₂ and 2F₂	Real, inverted, diminished
At 2F₁	At 2F₂	Real, inverted, same size
Between F₁ and 2F₁	Beyond 2F₂	Real, inverted, magnified
At F₁	At infinity	Real, inverted, highly magnified
Between F₁ and O	Same side as object	Virtual, erect, magnified

A concave lens always gives a virtual, erect, diminished image on the same side as the object: between F₁ and O, or at F₁ (point-sized) for an object at infinity.

Power and combinations

$P = 1/f$ (f in metres) in dioptres. For lenses in contact, powers simply add, which is how optometrists build up a prescription.

The human eye

The eye lens forms a real, inverted image on the retina, and the ciliary muscles change its focal length (accommodation). Myopia (far point closer than infinity) is corrected with a concave lens; hypermetropia (near point beyond 25 cm) with a convex lens; presbyopia is the age-related loss of accommodation (the near point moves away). A person with both near- and far-sightedness needs bifocal lenses: concave in the upper part, convex in the lower part.