Focal Length of Liquid Filled Lens Combination

❓ Question

A concavo-convex lens of refractive index:

$$\mu_g=\frac{3}{2}$$

and radii of curvature:

$$R_1=30\text{ cm}, \quad R_2=20\text{ cm}$$

is filled with a liquid of refractive index:

$$\mu_l=\frac{13}{10}$$

Find the focal length of the liquid-glass combination.

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Lens maker formula
👉 Combination of media
👉 Power of curved surfaces.

Main idea:

For lens filled with liquid:

$$P=P_1+P_2$$

where power of each surface depends on refractive indices across that surface.

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🔷 Step 1 — Identify Media 💯

Outside medium:

$$\mu=1$$

Glass refractive index:

$$\mu_g=\frac{3}{2}$$

Liquid refractive index:

$$\mu_l=\frac{13}{10}$$

Radii:

$$R_1=+30\text{ cm}$$
$$R_2=+20\text{ cm}$$

(Using sign convention for given shape.)

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🔷 Step 2 — Power of First Surface

For refraction at spherical surface:

$$P_1=\frac{\mu_2-\mu_1}{R}$$

First surface:

$$1 \rightarrow \frac{3}{2}$$

So:

$$P_1= \frac{\frac32-1}{30}$$
$$= \frac{1}{60}$$

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🔷 Step 3 — Power of Second Surface

Second surface separates:

$$\frac32 \rightarrow \frac{13}{10}$$

Hence:

$$P_2= \frac{\frac{13}{10}-\frac32}{20}$$
$$= \frac{13-15}{200}$$
$$= -\frac{1}{100}$$

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🔷 Step 4 — Total Power

$$P=P_1+P_2$$
$$= \frac1{60}-\frac1{100}$$

Taking LCM:

$$P=\frac{5-3}{300}$$
$$P=\frac1{150}$$

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🔷 Step 5 — Focal Length

$$f=\frac1P$$
$$f=150\text{ cm}$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ Normal lens maker formula directly use kar lena

❌ Second surface ke refractive indices ulte le lena

❌ Radius sign convention galat lena

Remember:

$$\boxed{ P=\sum \frac{\mu_2-\mu_1}{R} }$$

for each refracting surface separately.

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✅ Final Answer

$$\boxed{150\text{ cm}}$$

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