What is C if light AO=2î−3ĵ refracts to OB=Cî−4ĵ with μ₁=1, μ₂=1.5?

C≈1.6. sinθᵢ=2/√13, sinθᵣ=2/(1.5√13)≈0.37, C=4tanθᵣ≈1.6.

How to apply Snell's law using ray vectors?

sinθ = (parallel component)/(magnitude). For AO=(2,−3): sinθᵢ=2/√13. Apply μ₁sinθᵢ=μ₂sinθᵣ. Find C from refracted ray geometry.

Light incident along AO=2î−3ĵ emerges along OB=Cî−4ĵ μ₁=1 μ₂=1.5 — find C

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Light ray incident along $\overrightarrow{AO}=2\hat{i}-3\hat{j}$ emerges along $\overrightarrow{OB}=C\hat{i}-4\hat{j}$. ($\mu_1=1$, $\mu_2=1.5$). Find $C$:

A) 1.6 B) 0.16 C) 11.6 D) 16

✅ Correct Answer

A) 1.6

✓

Solution

Find sin θᵢ from incident ray

Interface normal is $\hat{j}$. Parallel component (x) relates to $\sin\theta$.

$|\overrightarrow{AO}|=\sqrt{4+9}=\sqrt{13}$

$\sin\theta_i=\dfrac{|x\text{-comp}|}{|AO|}=\dfrac{2}{\sqrt{13}}\approx0.5547$

Apply Snell's Law

$\mu_1\sin\theta_i=\mu_2\sin\theta_r$

$\sin\theta_r=\dfrac{2}{1.5\sqrt{13}}\approx0.3698$

$\cos\theta_r\approx0.929,\quad\tan\theta_r\approx0.398$

Find C from refracted ray

$C=|y\text{-comp of }OB|\times\tan\theta_r=4\times0.398\approx1.59\approx\boxed{1.6}$

📘 Snell's Law via Vector Components

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Important Concepts

Normal DirectionNormal to interface is ĵ. The x-component of the ray is parallel to the interface and determines sinθ.

sin θᵢ=2/√13|AO|=√(4+9)=√13. x-comp=2. sinθᵢ=2/√13≈0.5547.

Snell's Law1×(2/√13)=1.5×sinθᵣ → sinθᵣ=2/(1.5√13)≈0.3698.

C=4 tan θᵣ≈1.6OB has y-comp=4. C/4=tanθᵣ≈0.398. C≈1.59≈1.6.

FAQs

How to identify normal direction?

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Light travels in x-y plane. The interface is horizontal (boundary between media). Normal is ĵ (vertical). Components along ĵ are normal components; along î are tangential.

What is sinθᵣ?

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sinθᵣ=μ₁sinθᵢ/μ₂=0.5547/1.5≈0.3698.

Why C=4 tanθᵣ?

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OB=(C,−4). The normal direction is ĵ: |y-comp|=4 represents the 'cosθ' direction. tanθᵣ=C/4.

More precise C?

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101C²=256 → C=√(256/101)≈1.592≈1.6.

Is this from JEE Main 2026?

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Yes, this appeared in JEE Main 2026.

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