Q.What is the focal length of a convex lens of focal length $30\ \text{cm}$ in contact with a concave lens of focal length $20\ \text{cm}$? Is the system a converging or a diverging lens? Ignore thickness of the lenses.

The key idea is the effective focal length of two thin lenses in contact, given by the formula:

Step 1: Assign signs correctly. For a convex lens, $f_1 = +30\ \text{cm}$. For a concave lens, $f_2 = -20\ \text{cm}$.

Step 2: Substitute into the formula:

Step 3: Take LCM (60):

$\frac{1}{F} = \frac{2 - 3}{60} = -\frac{1}{60}$ …

The key to solving this lies in understanding how lenses combine when placed in contact. Instead of working directly with focal lengths, it's far easier to use power — the quantity that adds linearly.

A convex lens converges light — by convention, its focal length is positive. A concave lens diverges light — its focal length is negative. This sign convention is crucial.

Let's work through it step by step.

1. Write the focal lengths with correct signs.

   Convex lens: $f_1 = +30\ \text{cm} = +0.30\ \text{m}$

   Concave lens: $f_2 = -20\ \text{cm} = -0.20\ \text{m}$

2. Convert each to power.

   $P_1 = \frac{1}{+0.30} = +\frac{10}{3}\ \text{D} \approx +3.33\ \text{D}$

   $P_2 = \frac{1}{-0.20} = -5\ \text{D}$

3. Add the powers.

   $P_{\text{eq}} = P_1 + P_2 = \frac{10}{3} - 5 = \frac{10}{3} - \frac{15}{3} = -\frac{5}{3}\ \text{D}$

   The negative sign tells us the combination behaves like a diverging lens overall.

4. Convert back to focal length.

   $f_{\text{eq}} = \frac{1}{P_{\text{eq}}} = \frac{1}{-5/3} = -\frac{3}{5}\ \text{m} = -0.60\ \text{m} = -60\ \text{cm}$ …

Method: Power of a Lens System in Contact

When thin lenses are placed in contact, the total power of the system is the algebraic sum of their individual powers. This is the Power Addition Method.

Steps

Step 1: Recall the formula for power of a lens

Power $P$ (in dioptres) is given by:

Step 2: Convert focal lengths to metres

• Convex lens: $f_1 = +30\ \text{cm} = +0.30\ \text{m}$
• Concave lens: $f_2 = -20\ \text{cm} = -0.20\ \text{m}$

Sign convention: Convex = converging = positive focal length. Concave = diverging = negative focal length.

Step 3: Calculate individual powers

Step 4: Add powers to get total power

$P_{\text{total}} = P_1 + P_2 = \frac{10}{3} - 5 = \frac{10}{3} - \frac{15}{3} = -\frac{5}{3} \ \text{D}$ …

Here are the common mistakes students make on this exact problem, along with the concept-first reasoning to avoid each.

Mistake 1: Using the wrong sign for the concave lens focal length

The mistake:

Plugging $f_2 = +20\ \text{cm}$ into the lens combination formula.

Why it’s wrong:

By the Cartesian sign convention (used in all Indian board exams):

• Convex lens → $f$ is positive.
• Concave lens → $f$ is negative.

So for this problem:

• $f_1 = +30\ \text{cm}$ (convex)
• $f_2 = -20\ \text{cm}$ (concave)

How to avoid:

Always write the sign explicitly before substituting. Memorise:

Convex = positive, Concave = negative (for focal length).

Mistake 2: Adding focal lengths directly

The mistake:

Thinking the combined focal length $F$ is $f_1 + f_2$.

Why it’s wrong:

Lenses in contact do not add like resistors in series. The correct formula is:

This comes from the fact that the total power (in dioptres) adds:

Since $P = \frac{1}{f\ (\text{in metres})}$, the focal lengths combine via reciprocals.

How to avoid:

Write the formula every time before substituting numbers. Never shortcut to $F = f_1 + f_2$.

Mistake 3: Forgetting to convert units

The mistake:

Using cm directly in the power formula without converting to metres.

Why it’s not always wrong:

If you keep everything in cm, the formula $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$ still works as long as all $f$ values are in the same unit. The answer $F$ will also be in that unit.

But the trap:

If you later convert $F$ to power in dioptres, you must use metres. Many exam questions ask for power separately.

How to avoid:

• For focal length: work entirely in cm (or m), but stay consistent.
• For power: always convert to metres: $P = \frac{1}{f\ (\text{m})}$.

Mistake 4: Misinterpreting the sign of the final answer

The mistake:

Getting $F = +60\ \text{cm}$ and calling the system converging.

Why it’s wrong:

Let’s do the correct calculation:

So:

Negative focal length → the system behaves as a diverging lens.

How to avoid:

After calculating, check the sign:

• $F > 0$ → converging (convex-like)
• $F < 0$ → diverging (concave-like)

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