Numerical on Power of a Lens for CBSE Class 10

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Understanding the Power of a Lens: Concepts, Formulas, and Numericals

When studying optics, understanding how a lens bends light is crucial. Whether you are dealing with a simple magnifying glass or complex optical instruments, the ability of a lens to converge or diverge light rays is measured by its Power.

Here is a complete breakdown of the power of a lens, the sign conventions you need to remember, and how to solve standard numerical problems.

What is the Power of a Lens?

The power of a lens is defined as the measure of its ability to converge or diverge the light rays falling on it.

Note:
(i) A lens with a short focal length bends the light rays more, bringing them to a focus closer to the optical center, which means it has a higher power.
(ii) A lens with a longer focal length bends the light rays less, meaning it has a lower power.

Mathematically, the power of a lens is the reciprocal of its focal length.
$P = \frac{1}{f(\text{in meters})}$ and $P_{net} = P_1 + P_2$

SI Unit of Power


The SI unit of the power of a lens is the Diopter, denoted by the capital letter D.

Definition of 1 Diopter: One diopter is the power of a lens whose focal length is exactly 1 meter.

Sign Conventions for Lenses


Just like focal length, the power of a lens follows the standard Cartesian sign convention. This is highly important when solving exam numericals:
(i) Convex Lens (Converging Lens): A convex lens has a positive focal length. Therefore, its power is always positive (+).
(ii) Concave Lens (Diverging Lens): A concave lens has a negative focal length. Therefore, its power is always negative (-).

Numerical Problems on Power of Lenses

Class 10 Science · Light – Reflection and Refraction
N

Lens Power Practice Set 1

Apply the formulas $P = \frac{1}{f(\text{in meters})}$ and $P_{net} = P_1 + P_2$
Ch 10 · Light
1
A concave lens produces an image 20 cm from the lens of an object placed 30 cm from the lens. Calculate the power of the lens.
Answer & Solution

Answer: $-1.67\text{ D}$

Explanation: For a concave lens, the image is virtual and forms on the same side as the object. So, $u = -30\text{ cm}$ and $v = -20\text{ cm}$.

$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-20} – \left(-\frac{1}{30}\right) = -\frac{3}{60} + \frac{2}{60} = -\frac{1}{60}$$

Focal length $f = -60\text{ cm} = -0.6\text{ m}$.

$$P = \frac{1}{f} = \frac{1}{-0.6} = -1.67\text{ D}$$
2
A convex lens is of focal length 10 cm. What is its power?
Answer & Solution

Answer: $+10\text{ D}$

Explanation: The focal length of a convex lens is positive. $f = +10\text{ cm}$. First, convert it to meters: $f = 0.1\text{ m}$.

$$P = \frac{1}{f(\text{in m})} = \frac{1}{0.1} = +10\text{ D}$$
3
A person having a myopia eye uses a concave lens of focal length 50 cm. What is the power of the lens?
Answer & Solution

Answer: $-2\text{ D}$

Explanation: The focal length of a concave lens is negative. $f = -50\text{ cm} = -0.5\text{ m}$.

$$P = \frac{1}{f(\text{in m})} = \frac{1}{-0.5} = -2\text{ D}$$
4
A thin lens has a focal length of –25 cm. What is the power of the lens and what is its nature?
Answer & Solution

Answer: $P = -4\text{ D}$, Concave (Diverging) lens

Explanation: Since the focal length is negative, the lens is concave in nature. Convert $f$ to meters: $f = -25\text{ cm} = -0.25\text{ m}$.

$$P = \frac{1}{-0.25} = -4\text{ D}$$
5
A lens has a power of –2.5 D. What is the focal length and nature of the lens?
Answer & Solution

Answer: $f = -40\text{ cm}$, Concave lens

Explanation: The negative power indicates it is a concave (diverging) lens. $P = -2.5\text{ D}$.

$$f = \frac{1}{P} = \frac{1}{-2.5} = -0.4\text{ m}$$

Converting to centimeters: $-0.4\text{ m} = -40\text{ cm}$.

6
Find the power of a concave lens of focal length 2 m.
Answer & Solution

Answer: $-0.5\text{ D}$

Explanation: The focal length of a concave lens is negative, so $f = -2\text{ m}$.

$$P = \frac{1}{f} = \frac{1}{-2} = -0.5\text{ D}$$
7
A convex lens forms a real and inverted image of needle at a distance of 50 cm from the lens. If the image is of the same size as the needle, where is the needle placed in front of the lens? Also, find the power of the lens.
Answer & Solution

Answer: Needle is 50 cm in front of the lens. $P = +4\text{ D}$

Explanation: A convex lens forms a real, inverted image of the same size only when the object is placed at $2F$. Thus, the image is also formed at $2F$ on the other side.

Given $v = +50\text{ cm}$. Since $v = 2F \implies 2F = 50\text{ cm} \implies F = 25\text{ cm}$.

The object (needle) is placed at $-50\text{ cm}$ (in front of the lens).

Focal length in meters $f = +0.25\text{ m}$.

$$P = \frac{1}{f} = \frac{1}{0.25} = +4\text{ D}$$
8
Two thin lenses of power +3.5 D and –2.5 D are placed in contact. Find the power and focal length of the lens combination.
Answer & Solution

Answer: $P = +1.0\text{ D}$, $f = 100\text{ cm}$

Explanation: The power of a combination of lenses is the algebraic sum of their individual powers.

$$P = P_1 + P_2 = +3.5 + (-2.5) = +1.0\text{ D}$$

Focal length of the combination:

$$f = \frac{1}{P} = \frac{1}{+1.0} = +1\text{ m} = +100\text{ cm}$$
9
A doctor has prescribed a corrective lens of power –1.5 D. Find the focal length of the lens. Is the prescribed lens is diverging or converging?
Answer & Solution

Answer: $f = -66.67\text{ cm}$. It is a diverging lens.

Explanation: Given $P = -1.5\text{ D}$. The negative sign indicates a concave (diverging) lens.

$$f = \frac{1}{P} = \frac{1}{-1.5} = -0.6667\text{ m}$$

Converting to centimeters: $-66.67\text{ cm}$.

10
A concave lens of focal length 25 cm and a convex lens of focal length 20 cm are placed in contact with each other. What is the power of this combination? Also, calculate focal length of the combination.
Answer & Solution

Answer: $P = +1\text{ D}$, $f = +100\text{ cm}$

Explanation: First, find the powers of individual lenses.
Concave lens: $f_1 = -25\text{ cm} = -0.25\text{ m} \implies P_1 = \frac{1}{-0.25} = -4\text{ D}$
Convex lens: $f_2 = +20\text{ cm} = +0.20\text{ m} \implies P_2 = \frac{1}{0.20} = +5\text{ D}$

Power of combination:

$$P = P_1 + P_2 = -4 + 5 = +1\text{ D}$$

Focal length of combination:

$$f = \frac{1}{P} = \frac{1}{1} = +1\text{ m} = +100\text{ cm}$$
N

Lens Power Practice Set 2

Advanced combinations and image position calculations
Ch 10 · Light
11
A convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 10 cm. What is the focal length and power of the combination?
Answer & Solution

Answer: $P = -5\text{ D}$, $f = -20\text{ cm}$

Explanation:
Convex lens: $f_1 = +0.20\text{ m} \implies P_1 = +5\text{ D}$
Concave lens: $f_2 = -0.10\text{ m} \implies P_2 = -10\text{ D}$

$$P_{net} = 5 + (-10) = -5\text{ D}$$
$$f_{net} = \frac{1}{-5} = -0.2\text{ m} = -20\text{ cm}$$
12
An object is placed at a distance of 50 cm from a concave lens of focal length 30 cm. Find the nature and position of the image.
Answer & Solution

Answer: $v = -18.75\text{ cm}$. Image is virtual, erect and diminished.

Explanation: Given $u = -50\text{ cm}$ and $f = -30\text{ cm}$. Using the lens formula:

$$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-30} + \frac{1}{-50} = -\frac{5}{150} – \frac{3}{150} = -\frac{8}{150}$$
$$v = -\frac{150}{8} = -18.75\text{ cm}$$
13
An object of height 2 cm is placed at a distance of 15 cm in front of a concave lens of power –10 D. Find the size of the image.
Answer & Solution

Answer: $0.8\text{ cm}$

Explanation: Object height $h = 2\text{ cm}$, $u = -15\text{ cm}$. $P = -10\text{ D} \implies f = -0.1\text{ m} = -10\text{ cm}$.

$$\frac{1}{v} = \frac{1}{-10} + \frac{1}{-15} = -\frac{3}{30} – \frac{2}{30} = -\frac{5}{30} = -\frac{1}{6}$$

So, $v = -6\text{ cm}$. Now find magnification ($m$):

$$m = \frac{v}{u} = \frac{-6}{-15} = +0.4$$
$$h’ = m \times h = 0.4 \times 2 = 0.8\text{ cm}$$
14
A convergent lens of power 8 D is combined with a divergent lens of power –10 D. Calculate focal length of the combination.
Answer & Solution

Answer: $-50\text{ cm}$

Explanation: Given $P_1 = +8\text{ D}$ and $P_2 = -10\text{ D}$.

$$P_{net} = P_1 + P_2 = 8 + (-10) = -2\text{ D}$$
$$f = \frac{1}{P} = \frac{1}{-2} = -0.5\text{ m} = -50\text{ cm}$$
15
A concave lens is kept in contact with a convex lens of focal length 20 cm. The combination works as a converging lens of focal length 100 cm. Calculate power of concave lens.
Answer & Solution

Answer: $-4\text{ D}$

Explanation: Let the concave lens be lens 1.
Convex lens: $f_2 = +0.20\text{ m} \implies P_2 = +5\text{ D}$.
Combination: $f_{net} = +1.00\text{ m} \implies P_{net} = +1\text{ D}$.

$$P_{net} = P_1 + P_2 \implies 1 = P_1 + 5$$
$$P_1 = 1 – 5 = -4\text{ D}$$
16
Find the focal length and nature of lens which should be placed in contact with a lens of focal length 10 cm so that the power of the combination becomes 5 D.
Answer & Solution

Answer: $f = -20\text{ cm}$, Concave lens

Explanation: Given lens: $f_1 = +0.10\text{ m} \implies P_1 = +10\text{ D}$. Target power $P_{net} = +5\text{ D}$.

$$P_{net} = P_1 + P_2 \implies 5 = 10 + P_2 \implies P_2 = -5\text{ D}$$
$$f_2 = \frac{1}{P_2} = \frac{1}{-5} = -0.2\text{ m} = -20\text{ cm}$$

Since the focal length is negative, it must be a concave lens.

17
A convex lens of power 3 D is held in contact with a concave lens of power – 1 D. A parallel beam of light is made to fall on the combination. At what distance from the combination will the beam get focused?
Answer & Solution

Answer: $50\text{ cm}$

Explanation: A parallel beam of light always focuses at the principal focus of a lens system. Therefore, we just need to find the focal length.
$P_1 = +3\text{ D}$, $P_2 = -1\text{ D}$.

$$P_{net} = 3 – 1 = +2\text{ D}$$
$$f_{net} = \frac{1}{2} = +0.5\text{ m} = 50\text{ cm}$$
18
A convex lens of focal length 25 cm and a concave lens of focal length 10 cm are placed in close contact with one another.
a) What is the power of the combination?
b) What is the focal length of the combination?
c) Is this combination converging or diverging?
Answer & Solution

Answer: (a) $-6\text{ D}$, (b) $-16.67\text{ cm}$, (c) Diverging

Explanation:
Convex: $f_1 = +0.25\text{ m} \implies P_1 = +4\text{ D}$
Concave: $f_2 = -0.10\text{ m} \implies P_2 = -10\text{ D}$

(a) $P_{net} = 4 – 10 = -6\text{ D}$

(b) $f_{net} = \frac{1}{-6}\text{ m} = -0.1667\text{ m} = -16.67\text{ cm}$

(c) Because the net power/focal length is negative, the combination acts as a diverging lens.

19
The power of a combination of two lenses X and Y is 5 D. If the focal length of lens X be 15 cm, then
a) calculate the focal length of lens Y.
b) State the nature of the lens Y.
Answer & Solution

Answer: (a) $-60\text{ cm}$ (b) Concave (diverging)

Explanation: $P_{net} = +5\text{ D}$.
Lens X: $f_x = 0.15\text{ m} \implies P_x = \frac{1}{0.15} = \frac{100}{15} = +6.67\text{ D}$

$$P_{net} = P_x + P_y \implies 5 = 6.67 + P_y \implies P_y = -1.67\text{ D}$$

(a) $f_y = \frac{1}{-1.67} \approx -0.6\text{ m} = -60\text{ cm}$. (More precisely, $P_y = 5 – \frac{20}{3} = -\frac{5}{3}\text{ D}$, so $f_y = -\frac{3}{5}\text{ m} = -60\text{ cm}$).

(b) Since the focal length is negative, Lens Y is concave.

20
Two lenses A and B have focal lengths of +20 cm and –10 cm, respectively.
a) What is the nature of lens A and lens B?
b) What is the power of lens A and lens B?
c) What is the power of the combination if lenses A and B are held close together?
Answer & Solution

Answer: (a) A is Convex, B is Concave (b) $P_A = +5\text{ D}, P_B = -10\text{ D}$ (c) $P_{net} = -5\text{ D}$

Explanation:

(a) Positive focal length ($+20\text{ cm}$) means Lens A is Convex. Negative focal length ($-10\text{ cm}$) means Lens B is Concave.

(b) $P_A = \frac{1}{0.2} = +5\text{ D}$. $P_B = \frac{1}{-0.1} = -10\text{ D}$.

(c) $P_{net} = P_A + P_B = 5 – 10 = -5\text{ D}$.

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