Ray Optics

JEE Main level, single correct, medium

75 of 295

Question 75 of 295

Light ray incident along a vector AO⃗\vec{AO} (AO⃗=2i^−3j^)(\vec{AO} = 2\hat{i}-3\hat{j}) emerges out along vector OB⃗\vec{OB} (OB⃗=Ci^−4j^)(\vec{OB} = C\hat{i}-4\hat{j}) as shown in the figure below. The value of CC is ________.

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Answer: (A) 1.61.6

From the given figure, the interface is horizontal and the normal is along the y-axis.

The incident ray is along the vector AO⃗=2i^−3j^\vec{AO} = 2\hat{i} - 3\hat{j}.
The angle of incidence α\alpha is the angle between the incident ray and the normal.
sin⁡α=222+(−3)2=213\sin \alpha = \dfrac{2}{\sqrt{2^2 + (-3)^2}} = \dfrac{2}{\sqrt{13}}

The refracted ray is along the vector OB⃗=Ci^−4j^\vec{OB} = C\hat{i} - 4\hat{j}.
The angle of refraction β\beta is the angle between the refracted ray and the normal.
sin⁡β=CC2+(−4)2=CC2+16\sin \beta = \dfrac{C}{\sqrt{C^2 + (-4)^2}} = \dfrac{C}{\sqrt{C^2 + 16}}

Applying Snell's law at the interface:
μ1sin⁡α=μ2sin⁡β\mu_1 \sin \alpha = \mu_2 \sin \beta
1×213=1.5×CC2+161 \times \dfrac{2}{\sqrt{13}} = 1.5 \times \dfrac{C}{\sqrt{C^2 + 16}}
213=32CC2+16\dfrac{2}{\sqrt{13}} = \dfrac{3}{2} \dfrac{C}{\sqrt{C^2 + 16}}

Squaring both sides:
413=94(C2C2+16)\dfrac{4}{13} = \dfrac{9}{4} \left( \dfrac{C^2}{C^2 + 16} \right)
16(C2+16)=117C216(C^2 + 16) = 117 C^2
16C2+256=117C216 C^2 + 256 = 117 C^2
101C2=256101 C^2 = 256
C2=256101C^2 = \dfrac{256}{101}
C=16101≈1.592C = \dfrac{16}{\sqrt{101}} \approx 1.592

Approximating 101≈10\sqrt{101} \approx 10, we get C≈1.6C \approx 1.6.

Answer: 1.61.6

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