1/f = 1/f₁ + 1/f₂. Power of combination. Sign convention.

Combination of Thin Lenses in Contact

Question

Two thin convex lenses of focal lengths $f_1 = 20$ cm and $f_2 = 30$ cm are placed in contact coaxially. Find:

(a) The effective focal length of the combination (b) The power of the combination

(CBSE 2024 Board Exam)

Solution — Step by Step

Write the lens combination formula

For two thin lenses in contact, the effective focal length $f$ is given by:

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

This formula comes directly from adding the refractions at each lens surface — since the lenses share the same optical axis and are touching, the image from the first lens acts as the object for the second without any gap correction.

Substitute values (sign convention matters here)

Both lenses are convex, so both focal lengths are positive (using the standard sign convention where distances measured in the direction of incident light are positive).

$\frac{1}{f} = \frac{1}{20} + \frac{1}{30}$

Take LCM of 20 and 30, which is 60:

$\frac{1}{f} = \frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12}$

$\boxed{f = 12 \text{ cm}}$

Convert focal length to metres for Power

Power is defined as $P = \frac{1}{f}$ where $f$ must be in metres.

$f = 12 \text{ cm} = 0.12 \text{ m}$

$P = \frac{1}{0.12} = +8.33 \text{ D}$

Alternatively, use the direct addition of powers:

$P = P_1 + P_2 = \frac{100}{20} + \frac{100}{30} = 5 + 3.33 = \boxed{+8.33 \text{ D}}$

💡 Expert Tip

When focal length is in centimetres, Power in dioptres = $\frac{100}{f(\text{cm})}$ . This shortcut saves conversion time in board exams — you'll thank yourself in Part B questions.

Why This Works

Each lens bends light by a certain amount — its "convergence power." When two lenses sit in contact, their individual bending effects simply add up. The total deviation of the ray equals the sum of deviations from each lens.

Mathematically, if the first lens forms an image at distance $v_1$ , that image becomes the virtual object for the second lens at the same distance (since they're touching, there's no separation). Working through the geometry of both refractions gives us the addition formula $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$ naturally.

This is also why Power, not focal length, is the "natural" unit for opticians — the powers of lenses in a system simply add. Your optometrist writes –2.5 D on your prescription because it's the number that combines linearly.

Alternative Method — Using Power Addition Directly

Instead of working with focal lengths first, convert both to powers immediately:

$P_1 = \frac{100}{f_1(\text{cm})} = \frac{100}{20} = +5 \text{ D}$

$P_2 = \frac{100}{f_2(\text{cm})} = \frac{100}{30} = +3.\overline{3} \text{ D}$

$P = P_1 + P_2 = 5 + 3.\overline{3} = +8.\overline{3} \text{ D}$

Then recover focal length: $f = \frac{100}{P} = \frac{100}{8.\overline{3}} = 12$ cm.

This method is faster when the question asks for power first — start with what you need.

Common Mistake

⚠️ Common Mistake

The most common error here is ignoring the sign convention for a concave lens. If one of the lenses were concave with $f_2 = -30$ cm, you'd write $\frac{1}{f} = \frac{1}{20} + \frac{1}{-30}$ . Many students plug in 30 (positive) for both, get a more convergent system, and lose full marks. Always assign the sign before substituting — convex is positive, concave is negative.

A close second: writing focal length in centimetres when computing power. $P = \frac{1}{12}$ gives 0.083 D — a nonsensically weak lens. The formula $P = \frac{1}{f}$ demands metres. Memorise the shortcut $P = \frac{100}{f(\text{cm})}$ and skip the unit confusion entirely.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Power Addition Directly

Common Mistake

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