Is it possible for two lenses to produce zero power?

Conceptual Questions · Q3

Q. Is it possible for two lenses to produce zero power?

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Concept understanding — Power of a Lens and Combination of Lenses

The power of a lens is a measure of how strongly it converges or diverges light, defined as $P=1/f$ (SI unit dioptre, D; $1\,\text{D}=1\,\text{m}^{-1}$ ), positive for a converging lens and negative for a diverging one; from the lens maker's formula, $P=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$ , so a higher refractive index or more sharply curved (smaller-radius, 'bulkier') surfaces give greater power (and hence a shorter focal length), while gently curved ('skinny') lenses have low power. For two thin lenses of focal lengths $f_1,f_2$ placed in contact (a common optical centre), writing and adding the lens equation for each in turn (the first lens's image serving as the second lens's object) gives the effective focal length $\dfrac{1}{F}=\dfrac{1}{f_1}+\dfrac{1}{f_2}$ , extendable to any number of lenses as $\dfrac{1}{F}=\sum\dfrac1{f_i}$ , or equivalently as an algebraic sum of powers $P=P_1+P_2+\ldots$ (some positive for convex, some negative for concave); the total magnification of the combination is the product of the individual magnifications, $m=m_1\times m_2\times\cdots$ . Two thin lenses can even combine to give zero net power ( $P_1=-P_2$ ) if a converging and a diverging lens of exactly matching magnitudes are used together. When two thin lenses of focal length $f_1,f_2$ are separated by a distance $d$ and only parallel incident rays (an object at infinity) are considered, tracing the net deviation angle through both lenses gives the special-case formula $\dfrac{1}{F}=\dfrac{1}{f_1}+\dfrac{1}{f_2}-\dfrac{d}{f_1f_2}$ for the combined focal length, together with expressions locating the position of the single equivalent lens measured from each of the two real lenses; this out-of-contact formula is valid only for the special case of an object at infinity -- for an object at a finite distance the image must instead be found by applying the lens equation separately, lens by lens, in sequence.

Step 1. For two thin lenses of focal length $f_1,f_2$ placed in contact, the combined power is the algebraic sum of the individual powers, $P=P_1+P_2$ .

Step 2. If one lens is converging (positive power, $P_1>0$ ) and the other is diverging (negative power, $P_2<0$ ), and their magnitudes happen to be exactly equal, $P_2=-P_1$ , then $P=P_1+P_2=P_1-P_1=0$ .

Step 3. A combination of zero net power means $1/F=0$ , i.e. the equivalent focal length is infinite -- physically, the combination neither converges nor diverges parallel light at all, and behaves optically like a plain sheet of glass.

Step 4. So yes, it is entirely possible: for example, a converging lens of focal length $+20$ cm and a diverging lens of focal length $-20$ cm, placed in contact, combine to give exactly zero net power, $P=1/20+(-1/20)=0$ .

✓Final answer

Yes -- placing a converging lens and a diverging lens of exactly equal and opposite power in contact gives a combination with zero net power (infinite equivalent focal length), which passes light through with no net convergence or divergence.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 32 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. If power of a lens is +2 dioptre, then it is -$
(i) Concave lens of focal length 50 cm.
(ii) Convex lens of focal length 500 mm
(iii) Concave lens of focal length 0.25 m.
(iv) Convex lens of focal length 25 cm.
›Reveal solutionSolution

f = 1/P; positive power means a convex (converging) lens.

Power $P=+2$ D, so $f=\dfrac{1}{P}=\dfrac{1}{2}\text{ m}=0.5\text{ m}=50\text{ cm}=500\text{ mm}$ . Since the power is positive, the lens is converging, i.e. convex.

✓Final answer
$(ii) Convex lens of focal length 500 mm$
CBSE 2026Set ANNUAL1 markMCQ
$Q. Focal length of field lens of power 2D is:$
(a) 50 cm
(b) –50 cm
(c) 200 cm
(d) –200 cm
›Reveal solutionSolution

Focal length is the reciprocal of power; for $P = 2\,D$ , $f = 0.5\,\text{m} = 50\,\text{cm}$ .

Power of a lens is $P = \dfrac{1}{f(\text{in metres})}$ . Given $P = 2\,D$ : $f = \dfrac{1}{P} = \dfrac{1}{2} = 0.5\,\text{m} = 50\,\text{cm}$ . Since the power is positive, the lens is converging, and its focal length is positive.

✓Final answer
$Focal length = 50 cm (option a).$
CBSE 2025Set D1 markMCQ
$Q. If two converging lenses of equal focal length f are kept in contact then the focal length of the combination will be (A) f (B) 2f (C) f/2 (D) 3f$
›Reveal solutionSolution

For thin lenses in contact powers add: P = P₁ + P₂, giving F = f/2 for two equal lenses.

When two thin lenses are placed in contact, their combined power is the sum of the individual powers:

1/F = 1/f₁ + 1/f₂

With f₁ = f₂ = f:

1/F = 1/f + 1/f = 2/f ⟹ F = f/2

The combination is more converging (shorter focal length) than either lens alone.

✓Final answer
$(C) f/2.$
CBSE 2025Set D1 markMCQ
$Q. Power of a convex lens is 2 dioptre. Its focal length will be (A) 20 cm (B) 50 cm (C) 40 cm (D) 60 cm$
›Reveal solutionSolution

Focal length (in metres) is the reciprocal of power in dioptres: f = 1/P.

The power of a lens is P = 1/f, with f in metres. For P = 2 D:

f = 1/P = 1/2 = 0.5 m = 50 cm

Being a convex lens, f is positive. So the focal length is 50 cm.

✓Final answer
$(B) 50 cm.$
CBSE 2025Set A1 mark
$Q. Write answer in one sentence: What is power of a lens having +0.5 m focal length?$
›Reveal solutionSolution

A lens with focal length +0.5 m has power +2 dioptre.

The power of a lens is defined as the reciprocal of its focal length expressed in metres:

$P = \dfrac{1}{f(\text{in metres})}$

Here $f = +0.5$ m (positive, by sign convention for a converging/convex lens):

$P = \dfrac{1}{0.5} = 2\ \text{D (dioptre)}$

The positive sign confirms it is a converging lens; a diverging lens would have negative power.

✓Final answer
$P = +2 D (dioptre).$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Two thin lenses of focal length f1 and f2 are placed in contact with each other, the effective focal length of the combination will be$
(a) f = f1 + f2
(b) f = f1 - f2
(c) f = (f1 + f2) / (f1 f2)
(d) f = (f1 f2) / (f1 + f2)
›Reveal solutionSolution

For thin lenses in contact, the powers (reciprocals of focal length) simply add, so the combined focal length is the product over the sum of the individual focal lengths.

For a single thin lens, the lens maker's relation for an object gives $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ . For two thin lenses of focal length $f_1, f_2$ placed in contact, the image formed by the first lens acts as the object for the second. Adding the two lens equations for the combination:

$\dfrac{1}{v}-\dfrac{1}{u} = \left(\dfrac{1}{v_1}-\dfrac{1}{u}\right)+\left(\dfrac{1}{v}-\dfrac{1}{v_1}\right) = \dfrac{1}{f_1}+\dfrac{1}{f_2}$

So if $f$ is the equivalent focal length: $\dfrac{1}{f} = \dfrac{1}{f_1}+\dfrac{1}{f_2} = \dfrac{f_1+f_2}{f_1 f_2}$

$\Rightarrow f = \dfrac{f_1 f_2}{f_1+f_2}$

✓Final answer
$(d) f = (f1 f2)/(f1 + f2).$
CBSE 2025Set ANNUAL1 markMCQ
$Q. If the focal length of a lens is f metre, then the value of its power will be$
(a) f dioptre
(b) 1/f dioptre
(c) (1 - f) dioptre
(d) 100/f dioptre
›Reveal solutionSolution

The power of a lens is defined as P = 1/f, with f measured in metres, giving power in dioptres (D).

By definition, lens power P (in dioptres) is the reciprocal of the focal length f (in metres):

P = 1/f

A converging lens has positive power, a diverging lens negative power, and a shorter focal length means a stronger (higher-power) lens. Since f is already stated in metres here, the power is simply 1/f dioptre (no extra factor of 100 is needed - that factor only appears if f were given in centimetres).

✓Final answer
$(b) 1/f dioptre.$
CBSE 2025Set ANNUAL1 mark
$Q. Two thin lenses of power + 4D and - 2D are in contact. The focal length of the combination is ______.$
›Reveal solutionSolution

Powers of thin lenses in contact simply add; here P = (+4 D) + (−2 D) = +2 D, giving f = 1/P = 0.5 m.

For two thin lenses of powers $P_1$ and $P_2$ placed in contact (coaxially, touching), the power of the combination is the algebraic sum:

$P = P_1 + P_2$

Here $P_1 = +4\ \text{D}$ and $P_2 = -2\ \text{D}$ , so:

$P = 4 + (-2) = +2\ \text{D}$

Since power and focal length are related by $P = \dfrac{1}{f}$ (f in metres):

$f = \dfrac{1}{P} = \dfrac{1}{2} = 0.5\ \text{m} = 50\ \text{cm}$

The positive sign shows the combination behaves as a converging (convex) lens overall.

✓Final answer
$P = +2 D, so f = 0.5 m = 50 cm (converging).$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The tangent of the angle by which it converges or diverges a beam of light parallel to the principle axis falling at unit distance from optical centre is called:$
(a) Malus law
(b) Lens formula
(c) Power of a lens
(d) Snell's law.
›Reveal solutionSolution

The power of a lens is defined exactly as the tangent of the angle of convergence/divergence produced in a ray at unit distance from the optical centre.

If a parallel beam of light, on refraction through a lens, converges or diverges through an angle $\theta$ at unit distance from the optical centre, the power of the lens is $P = \tan\theta$ , which for small angles equals $1/f$ (f in metres), measured in dioptres (D). A convex (converging) lens has positive power, a concave (diverging) lens has negative power. This is distinct from Malus's law (polarisation), the lens formula ( $1/v - 1/u = 1/f$ ), and Snell's law (refraction at an interface).

✓Final answer
$(c) Power of a lens.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Two thin lenses of focal length f1 and f2 are in contact and coaxial. The power of the combination is -$
(a) (f1 + f2)/2
(b) (f1 + f2)/(f1 f2)
(c) sqrt(f1/f2)
(d) sqrt(f2/f1)
›Reveal solutionSolution

For thin lenses in contact, powers simply add.

For two thin lenses of focal lengths $f_1$ and $f_2$ placed in contact and coaxially, the powers add algebraically:

$P = P_1 + P_2 = \dfrac{1}{f_1} + \dfrac{1}{f_2} = \dfrac{f_1 + f_2}{f_1 f_2}$

✓Final answer
$P = \dfrac{f_1+f_2}{f_1 f_2}$
CBSE 2024Set IMPROVEMENT1 mark
$+2\ \text{D}$
›Reveal solutionSolution

Focal length is the reciprocal of power: $f = 1/P$ .

The power of a lens is related to its focal length (in metres) by $P=\dfrac{1}{f}$ . Given $P=+2\ \text{D}$ :

$f=\frac{1}{P}=\frac{1}{2}=0.5\ \text{m}=50\ \text{cm}$

Since the power is positive, the lens is converging (convex), consistent with the given convex lens.

✓Final answer
$f = 0.5\ \text{m} = 50\ \text{cm}$
CBSE 2024Set A1 markMCQ
$Q. Powers of two lenses kept in contact, are P₁ and P₂. The power of equivalent lens will be (A) P₁/P₂ (B) P₂/P₁ (C) P₁ \times P₂ (D) P₁ + P₂$
›Reveal solutionSolution

Powers of thin lenses in contact add up: P = P₁ + P₂.

For two thin lenses of focal lengths $f_1$ and $f_2$ placed in contact, the equivalent focal length is

$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}.$

Since power is the reciprocal of focal length, $P = 1/f$ (with f in metres), this becomes

$P = P_1 + P_2.$

So the powers simply add. (This additivity is why power, rather than focal length, is the convenient quantity for combining lenses, e.g. in spectacles.)

✓Final answer
$(D) P₁ + P₂.$

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Q.Is it possible for two lenses to produce zero power?

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