Lens in Water Trick in 59 Seconds! 🔥 | JEE Physics

❓ Question

A lens having refractive index $1.6$ has a focal length of 12 cm when it is placed in air.
Find the focal length of the same lens when it is placed in water, given that the refractive index of water is 1.28.

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🖼️ Question Image

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✍️ Short Solution

This is a pure concept-based JEE optics question.

👉 The shape of the lens does NOT change,
👉 Only the surrounding medium changes,
👉 So we compare focal lengths using the lens maker idea in different media.

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🔹 Step 1 — Key lens concept (VERY IMPORTANT 🔥)**

For a thin lens:

$$\frac{1}{f}\propto (\frac{{\mu }_{\text{lens}}}{{\mu }_{\text{medium}}}-1)$$

Where:

• $\mu_{\text{lens}}$ = refractive index of lens material

• $\mu_{\text{medium}}$ = refractive index of surrounding medium

📌 Since radius of curvature is same, we compare focal lengths by ratio only.

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🔹 Step 2 — Write ratio formula

Let:

• $f_a$ = focal length in air

• $f_w$ = focal length in water

Then:

$$\frac{f_w}{f_a}=\frac{(\frac{{\mu }_{\text{lens}}}{{\mu }_{\text{air}}}-1)}{(\frac{{\mu }_{\text{lens}}}{{\mu }_{\text{water}}}-1)}$$

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🔹 Step 3 — Substitute given values

Given:

$$\mu_{\text{lens}}=1.6, \quad \mu_{\text{air}}=1, \quad \mu_{\text{water}}=1.28, \quad f_a=12\text{ cm}$$

Compute numerator:

$$\frac{1.6}{1}-1=0.6$$

Compute denominator:

$$\frac{1.6}{1.28}-1 = 1.25-1 = 0.25$$

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🔹 Step 4 — Find focal length in water

$$\frac{f_w}{12} = \frac{0.6}{0.25} = 2.4$$

So:

$$f_w=12\times2.4=28.8\text{ cm}$$

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🔹 Step 5 — Physical interpretation (JEE insight 🧠)**

• In water, lens power decreases

• Reason: refractive index difference reduces

• Hence focal length increases

📌 Agar medium ka refractive index lens ke paas ho jaaye,
toh lens almost weak ho jaata hai.

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✅ Final Answer

$$\boxed{28.8\text{ cm}}$$

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