Power of a Lens

• The relationship between power and focal length is an inverse proportionality. As the focal length ($f$) decreases, the power ($P$) increases, reflecting the lens's increased ability to bend light over a shorter distance.

• Mathematically, this is expressed as: $P = \frac{1}{f}$ where $P$ is in dioptres (D) and $f$ is in meters (m).

• For converging lenses (convex), the focal length is considered positive, resulting in a positive power value. Conversely, diverging lenses (concave) have a negative focal length and thus a negative power value.

Calculating Power from Focal Length

• Step 1: Identify the focal length of the lens. If it is given in centimeters or millimeters, it must be converted to meters (e.g., $20\text{ cm} = 0.2\text{ m}$).
• Step 2: Determine the lens type. Assign a positive sign for convex lenses and a negative sign for concave lenses.
• Step 3: Apply the formula $P = 1/f$ to find the power in dioptres.

Finding Focal Length from Power

• To find the focal length when power is known, rearrange the formula to $f = 1/P$. The resulting value will be in meters, which can then be converted to other units if required.

• When two or more thin lenses are placed in close contact (series), their combined effect can be treated as a single compound lens.

• The total power ($P_{total}$) of the combination is the algebraic sum of the individual powers of each lens: $P_{total} = P_1 + P_2 + P_3 + ... + P_n$

• This additive property allows for the creation of complex optical systems where the strengths of different lenses (both positive and negative) are combined to achieve a specific total refractive power.

• The Unit Trap: The most common error in exams is failing to convert the focal length from centimeters to meters. Always check your units before calculating $1/f$.

• Sign Consistency: Ensure that you apply the correct sign ($+$ or $-$) based on the lens type. A diverging lens must always have a negative power in your final answer.

• Reciprocal Check: When calculating $f$ from $P$, remember that a high power (e.g., $20\text{ D}$) must result in a very small focal length ($0.05\text{ m}$ or $5\text{ cm}$). If your answer is large, you likely forgot the reciprocal step.

• Combination Problems: For lenses in series, sum the powers algebraically. If a $+5\text{ D}$ lens is paired with a $-2\text{ D}$ lens, the total power is $+3\text{ D}$.