Refractive Index Layer Problem ⚡ JEE Shortcut

 

❓ Question

Container contains:

Liquid 1:
Refractive index $\mu_1 = 1.2$
Height = 60 cm

Liquid 2:
Refractive index $\mu_2 = 1.6$
Height = $H$

Viewed from above, apparent shift of bottom = 40 cm

Find $H$.

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🖼 Question Image

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✍️ Short Concept

For multiple layers:

$$\text{Apparent depth} = \sum \frac{\text{real depth}}{\mu}$$

Total apparent shift:

$$\text{Shift} = \text{Real depth} - \text{Apparent depth}$$

This is the MOST IMPORTANT relation.

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🔷 Step 1 — Write Real Depth 💯

Total real depth:

$$= 60 + H$$

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🔷 Step 2 — Write Apparent Depth

$$\text{Apparent depth} = \frac{60}{1.2} + \frac{H}{1.6}$$
$$= 50 + \frac{H}{1.6}$$

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🔷 Step 3 — Use Shift Formula

Given shift = 40 cm

$$\text{Shift} = \text{Real depth} - \text{Apparent depth}$$
$$40 = (60 + H) - \left(50 + \frac{H}{1.6}\right)$$

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🔷 Step 4 — Solve Equation

$$40 = 60 + H - 50 - \frac{H}{1.6}$$
$$40 = 10 + H - \frac{H}{1.6}$$
$$30 = H - \frac{H}{1.6}$$
$$30 = H\left(1 - \frac{1}{1.6}\right)$$
$$30 = H(1 - 0.625)$$
$$30 = 0.375H$$
$$H = 80 \text{ cm}$$

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✅ Final Answer

$$\boxed{H = 80 \text{ cm}}$$

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⭐ Golden JEE Insight

For layered liquids:

👉 Don’t use simple $\frac{d}{\mu}$ once.
👉 Apply separately for each layer.

Optical path length add hota hai —
depth directly add nahi hoti.