JEE Main Optics: Equivalent Power of Separated Lenses 💡

 

❓ Question

Two thin convex lenses of focal lengths

$$f_1 = 30\text{ cm}, \quad f_2 = 10\text{ cm}$$

are placed coaxially, 10 cm apart.

Find the power of the combination.

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🖼️ Question Image

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✍️ Short Solution

This is NOT the simple “lenses in contact” case.
Because they are separated by distance $d$,
we must use the correct combination formula 🔥

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🔹 Step 1 — Formula for Two Lenses Separated by Distance

Equivalent focal length:

$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$$

Where:

• $f_1, f_2$ in same units

• $d$ = separation

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🔹 Step 2 — Convert Everything to Same Unit

Given:

$$f_1 = 30\text{ cm},\quad f_2 = 10\text{ cm},\quad d = 10\text{ cm}$$

All already in cm — good 👍

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🔹 Step 3 — Substitute Values

$$\frac{1}{F} = \frac{1}{30} + \frac{1}{10} - \frac{10}{30\times 10}$$

Calculate step-by-step:

$$\frac{1}{30} + \frac{1}{10} = \frac{1}{30} + \frac{3}{30} = \frac{4}{30} = \frac{2}{15}$$

Now subtract third term:

$$\frac{10}{300} = \frac{1}{30}$$

So:

$$\frac{1}{F} = \frac{2}{15} - \frac{1}{30}$$

Convert to common denominator:

$$\frac{4}{30} - \frac{1}{30} = \frac{3}{30} = \frac{1}{10}$$

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🔹 Step 4 — Equivalent Focal Length

$$F = 10\text{ cm}$$

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🔹 Step 5 — Convert to Power

Power formula:

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

Convert 10 cm to meters:

$$10\text{ cm} = 0.1\text{ m}$$

So:

$$P = \frac{1}{0.1} = 10\text{ D}$$

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✅ Final Answer

$$\boxed{P = 10\ \text{Dioptre}}$$

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⭐ Golden JEE Insight

For two separated lenses:

$$P_{\text{total}} = P_1 + P_2 - d P_1 P_2$$

(where $d$ in meters)

🧠 Shortcut method:

$$P_1 = \frac{1}{0.3} = 3.33$$
$$P_2 = \frac{1}{0.1} = 10$$
$$P = 3.33 + 10 - (0.1)(3.33)(10)$$

Same result = 10 D