Lens Formula, Magnification, and Power of a Lens for CBSE Class 10

When an optometrist tests your eyes and writes a prescription such as "$-1.5\text{ D}"* or *"$+2.0\text{ D}", what do these positive and negative numbers mean? How does an optical lens manufacturer know the exact curvature required to focus light onto a camera sensor or onto the human retina?

In CBSE Class 10 Science, Chapter 9 (Light - Reflection and Refraction) concludes with the quantitative mathematics of lenses: the New Cartesian Sign Convention for Lenses, the Lens Formula, the Magnification Equation, and the concept of Power of a Lens.

What You Will Learn

• Cartesian Sign Convention applied to convex and concave lenses
• The Lens Formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (and why it has a minus sign)
• The Linear Magnification Formula: $m = \frac{h'}{h} = +\frac{v}{u}$
• Comparison: Mirror Formula vs. Lens Formula
• Definition, formula, and SI unit of the Power of a Lens (Dioptre)
• Power of combination of thin lenses in contact ($P = P_1 + P_2 + \dots$)
• Step-by-step solved numerical board exam problems

1. Cartesian Sign Convention for Lenses

Distances are measured using the Optical Center ($O$) as the origin $(0, 0)$:

1. Light travels from left to right; object is placed to the left of the lens.
2. Distances measured to the right of $O$ are positive ($+$).
3. Distances measured to the left of $O$ are negative ($-$).
4. Heights above the principal axis are positive ($+$); heights below are negative ($-$).

The Golden Focal Rules for Lenses:

1. Focal Length of a Convex Lens ($f$): Always POSITIVE ($+$) (converges to $F_2$ on the right).
2. Focal Length of a Concave Lens ($f$): Always NEGATIVE ($-$) (diverges from $F_1$ on the left).
3. Object Distance ($u$): Always NEGATIVE ($-$) for all lenses.

2. The Lens Formula

Statement

The mathematical relationship connecting object distance ($u$), image distance ($v$), and focal length ($f$) of a spherical lens is called the Lens Formula:

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

3. Linear Magnification ($m$) for Lenses

The ratio of the image height to the object height:

$\mathbf{m = \frac{\text{Height of Image } (h')}{\text{Height of Object } (h)} = +\frac{v}{u}}$

Crucial Comparison: Mirrors vs. Lenses

4. Power of a Lens ($P$)

The degree of convergence or divergence of light rays achieved by a lens depends directly on its focal length:

• A lens of short focal length bends light rays through large angles, focusing them close to the optical center.
• A lens of long focal length bends light rays gently, focusing them far away.

Formal Definition

The Power of a Lens is defined as the reciprocal of its focal length expressed in metres. It measures the ability of a lens to converge or diverge light rays.

Mathematical Formula:

$\mathbf{P = \frac{1}{f \text{ (in metres)}}} \quad \Longleftrightarrow \quad \mathbf{P = \frac{100}{f \text{ (in cm)}}}$

SI Unit of Power: The Dioptre (D)

• The SI unit of lens power is the Dioptre, denoted by the symbol $\text{D}$.
• Definition of $1\text{ Dioptre}$: <u>One Dioptre ($1\text{ D}$) is the power of a lens whose focal length is exactly $1\text{ metre}$ ($1\text{ D} = 1\text{ m}^{-1}$).</u>

Sign of Lens Power:

• Convex Lens: $f$ is positive $\implies$ Power is POSITIVE ($+$). (e.g., $+2.5\text{ D}$).
• Concave Lens: $f$ is negative $\implies$ Power is NEGATIVE ($-$). (e.g., $-1.5\text{ D}$).

5. Combination of Lenses in Contact

When multiple thin lenses are placed in direct contact:

1. Net Power is the Algebraic Sum: $\mathbf{P = P_1 + P_2 + P_3 + \dots}$
2. Net Focal Length: $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} + \dots$
3. Net Magnification is the Product: $\mathbf{m = m_1 \times m_2 \times m_3 \times \dots}$

Example: If a convex lens of power $+3.0\text{ D}$ is placed in contact with a concave lens of power $-1.0\text{ D}$, the net combination power is: $P = +3.0 + (-1.0) = \mathbf{+2.0\text{ D}}$ The combination acts as a converging lens of focal length $f = \frac{100}{2} = 50\text{ cm}$.

6. Solved CBSE Board Examination Problems

Solved Example 1: Concave Lens Problem (NCERT Classic)

Problem: A concave lens has focal length of $15\text{ cm}$. At what distance should the object from the lens be placed so that it forms an image at $10\text{ cm}$ from the lens? Also, find the magnification produced by the lens.

Solution:

1. Assign Cartesian Signs:
   • Concave lens focal length $f = \mathbf{-15\text{ cm}}$.
   • Concave lens always forms a virtual image on the same side $\implies v = \mathbf{-10\text{ cm}}$.
   • Object distance $u = ?$ and Magnification $m = ?$
2. Apply the Lens Formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{u} = \frac{1}{v} - \frac{1}{f}$ $\frac{1}{u} = \frac{1}{-10} - \left(\frac{1}{-15}\right) = -\frac{1}{10} + \frac{1}{15}$ LCM of 10 and 15 is 30: $\frac{1}{u} = \frac{-3 + 2}{30} = -\frac{1}{30} \implies \mathbf{u = -30\text{ cm}}$
3. Calculate Magnification ($m$): $m = \frac{v}{u} = \frac{-10}{-30} = \mathbf{+\frac{1}{3} = +0.33}$
4. Conclusion:
   • The object should be placed at a distance of $30\text{ cm}$ in front of the lens.
   • The positive sign of $m$ confirms the image is Virtual and Erect.
   • The value $0.33$ shows the image is diminished to one-third of the object's size.

Solved Example 2: Finding Power of a Lens

Problem: A doctor has prescribed a corrective lens of power $+1.5\text{ D}$. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Solution:

1. Power $P = +1.5\text{ D}$.
2. Formula: $P = \frac{1}{f\text{ (in metres)}} \implies f = \frac{1}{P}$: $f = \frac{1}{+1.5} = \frac{10}{15} = \frac{2}{3}\text{ m} = +0.67\text{ m} = \mathbf{+66.7\text{ cm}}$
3. Since the power and focal length are positive, <u>the prescribed lens is a convex (converging) lens</u> used to correct hypermetropia (farsightedness).

7. Summary and Examination Tips

Exam Tip: When calculating power ($P = 1/f$), students frequently forget to convert focal length from centimeters to metres! If $f = 20\text{ cm}$, writing $P = 1/20$ is completely wrong. You must write $P = 100/20 = +5\text{ D}$!

Common Mistake: Confusing the minus sign between mirror and lens formulas. Remember the mnemonic: Mirror has a Plus ($\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$), and Lens has a Less/minus ($\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$)!