(a) What is meant by 'power of a lens' ? (b) State and define the S.I. unit of power of a lens. (c) A convex lens of focal length 25 cm and a concave lens of focal length 10 cm are placed in close contact with each other. Calculate the lens power of this combination. OR (a) Draw a ray diagram to show the formation of image of an object placed between infinity and the optical centre of a concave lens. (b) A concave lens of focal length 15 cm forms an image 10 cm from the lens. Calculate (i) the distance of the object from the lens (ii) the magnification for the image formed (iii) the nature of the image formed

(a) Power of a lens:

The power of a lens is a measure of the degree of convergence or divergence of light rays falling on it. Mathematically, it is defined as the reciprocal of its focal length ($f$) expressed in meters.

$$P = \frac{1}{f \text{ (in meters)}}$$

(b) S.I. Unit of Power:

The S.I. unit of power of a lens is Dioptre, represented by the letter $D$.

Definition of 1 Dioptre: One dioptre is the power of a lens whose focal length is $1\text{ meter}$ ($1\text{ D} = 1\text{ m}^{-1}$). The power of a converging (convex) lens is positive, while that of a diverging (concave) lens is negative.

(c) Numerical Calculation:

• For the convex lens: $f_1 = +25\text{ cm} = +0.25\text{ m}$

$$P_1 = \frac{1}{f_1} = \frac{1}{+0.25\text{ m}} = +4\text{ D}$$

• For the concave lens: $f_2 = -10\text{ cm} = -0.10\text{ m}$

$$P_2 = \frac{1}{f_2} = \frac{1}{-0.10\text{ m}} = -10\text{ D}$$

• Net power of the combination ($P$):

$$P = P_1 + P_2 = +4\text{ D} + (-10\text{ D}) = -6\text{ D}$$

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OR

(a) Ray Diagram for Concave Lens:

When an object is placed between infinity and the optical centre ($O$) of a concave lens, the image is formed between the principal focus ($F_1$) and the optical centre ($O$) on the same side of the lens as the object. The image is virtual, erect, and diminished.

``

Diverging Ray

A | \

| \

| \

------B----F1----O---------F2------ (Principal Axis)

| / /

A'| / /

| / /

(Virtual Image A'B')

``

(b) Numerical Calculation:

Given for the concave lens:

• Focal length, $f = -15\text{ cm}$
• Image distance, $v = -10\text{ cm}$ (since a concave lens always forms a virtual image on the same side as the object)

(i) Object Distance ($u$):

Using the lens formula:

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

$$\frac{1}{-10} - \frac{1}{u} = \frac{1}{-15}$$

$$-\frac{1}{10} + \frac{1}{15} = \frac{1}{u}$$

$$\frac{-3 + 2}{30} = \frac{1}{u}$$

$$-\frac{1}{30} = \frac{1}{u} \implies u = -30\text{ cm}$$

Thus, the object is placed at a distance of $30\text{ cm}$ in front of the lens.

(ii) Magnification ($m$):

$$m = \frac{v}{u} = \frac{-10\text{ cm}}{-30\text{ cm}} = +\frac{1}{3} \approx +0.33$$

(iii) Nature of the Image:

Since the magnification is positive and its absolute value is less than $1$, the image is virtual, erect, and diminished.