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Geometrical Optics question

2019 · 12 Apr · Shift 2 · Q58
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  5. /2019 · 12 Apr · Shift 2 · Q58

Geometrical Optics question

2019 · 12 Apr · Shift 2 · Q58

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
A transparent cube of side, made of a material of refractive index μ\muμ 2, is immersed in a liquid of refractive index μ\muμ 1(μ\muμ 1 < μ\muμ 2). A ray is incident on the face AB at an angle θ\thetaθ(shown in the figure). Total internal reflection takes place at point E on the face BC. JEE Main 2019 (Online) 12th April Evening Slot Physics - Geometrical Optics Question 177 English Then θ\thetaθ must satisfy :
  1. A
    θ>sin⁡−1μ22μ12−1\theta \gt {\sin ^{ - 1}}\sqrt {{{\mu _2^2} \over {\mu _1^2}} - 1}θ>sin−1μ12​μ22​​−1​
  2. B
    θ<sin⁡−1μ22μ12−1\theta \lt {\sin ^{ - 1}}\sqrt {{{\mu _2^2} \over {\mu _1^2}} - 1}θ<sin−1μ12​μ22​​−1​
  3. C
    θ<sin⁡−1μ1μ2\theta \lt {\sin ^{ - 1}}{{{\mu _1}} \over {{\mu _2}}}θ<sin−1μ2​μ1​​
  4. D
    θ>sin⁡−1μ1μ2\theta \gt {\sin ^{ - 1}}{{{\mu _1}} \over {{\mu _2}}}θ>sin−1μ2​μ1​​
View written solutionFree

Correct answer: B

  1. Interpret the condition

A cube of refractive index μ2\mu_2μ2​ is immersed in a liquid of refractive index μ1\mu_1μ1​, with μ2>μ1\mu_2 > \mu_1μ2​>μ1​.

A ray is incident on one face of the cube at angle θ\thetaθ (with the normal). It enters the cube and then strikes an adjacent face. We want the condition on θ\thetaθ such that the ray does not emerge from that adjacent face, i.e. it undergoes total internal reflection there.

  1. Refraction at the first face

Let the refracted angle inside the cube at the first face be rrr.

By Snell's law,

μ1sin⁡θ=μ2sin⁡r\mu_1 \sin\theta = \mu_2 \sin rμ1​sinθ=μ2​sinr

so,

sin⁡r=μ1μ2sin⁡θ\sin r = \frac{\mu_1}{\mu_2}\sin\thetasinr=μ2​μ1​​sinθ
  1. Angle of incidence at the second face

Since the two faces of the cube are perpendicular, the angle of incidence on the adjacent face is

i=90∘−ri = 90^\circ - ri=90∘−r

For total internal reflection at the cube-liquid interface,

i>ici > i_ci>ic​

where the critical angle ici_cic​ satisfies

sin⁡ic=μ1μ2\sin i_c = \frac{\mu_1}{\mu_2}sinic​=μ2​μ1​​

Thus,

90∘−r>ic90^\circ - r > i_c90∘−r>ic​ r<90∘−icr < 90^\circ - i_cr<90∘−ic​

Taking sine on both sides,

sin⁡r<sin⁡(90∘−ic)=cos⁡ic\sin r < \sin(90^\circ - i_c) = \cos i_csinr<sin(90∘−ic​)=cosic​

Now,

cos⁡ic=1−sin⁡2ic=1−(μ1μ2)2=μ22−μ12μ2\cos i_c = \sqrt{1-\sin^2 i_c} = \sqrt{1-\left(\frac{\mu_1}{\mu_2}\right)^2} = \frac{\sqrt{\mu_2^2-\mu_1^2}}{\mu_2}cosic​=1−sin2ic​​=1−(μ2​μ1​​)2​=μ2​μ22​−μ12​​​

So the TIR condition becomes

sin⁡r<μ22−μ12μ2\sin r < \frac{\sqrt{\mu_2^2-\mu_1^2}}{\mu_2}sinr<μ2​μ22​−μ12​​​

Using sin⁡r=μ1μ2sin⁡θ\sin r = \dfrac{\mu_1}{\mu_2}\sin\thetasinr=μ2​μ1​​sinθ,

μ1μ2sin⁡θ<μ22−μ12μ2\frac{\mu_1}{\mu_2}\sin\theta < \frac{\sqrt{\mu_2^2-\mu_1^2}}{\mu_2}μ2​μ1​​sinθ<μ2​μ22​−μ12​​​ μ1sin⁡θ<μ22−μ12\mu_1 \sin\theta < \sqrt{\mu_2^2-\mu_1^2}μ1​sinθ<μ22​−μ12​​ sin⁡θ<μ22μ12−1\sin\theta < \sqrt{\frac{\mu_2^2}{\mu_1^2}-1}sinθ<μ12​μ22​​−1​

Therefore,

θ<sin⁡−1μ22μ12−1\theta < \sin^{-1}\sqrt{\frac{\mu_2^2}{\mu_1^2}-1}θ<sin−1μ12​μ22​​−1​

This matches Option B.

  1. Check options
  • A: opposite inequality, so incorrect.
  • B: correct.
  • C, D: these use sin⁡−1(μ1/μ2)\sin^{-1}(\mu_1/\mu_2)sin−1(μ1​/μ2​), which is the critical angle itself, not the required relation for the external incident angle θ\thetaθ.
  1. Final answer

The correct condition is

θ<sin⁡−1μ22μ12−1\boxed{\theta < \sin^{-1}\sqrt{\frac{\mu_2^2}{\mu_1^2}-1}}θ<sin−1μ12​μ22​​−1​​

So the correct option is B.

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