Optics 2 Question 1

1. A transparent cube of side d, made of a material of refractive index μ2, is immersed in a liquid of refractive index μ1(μ1<μ2). A ray is incident on the face AB at an angle θ (shown in the figure). Total internal reflection takes place at point E on the face BC.

Then, θ must satisfy

(2019 Main, 12 April II)

(a) θ<sin1μ1μ2

(b) θ>sin1μ22μ121

(c) θ<sin1μ22μ121

(d) θ>sin1μ1μ2

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Answer:

Correct Answer: 1. (c)

Solution:

  1. Key Idea The critical angle is defined as the angle of incidence that provides an angle of refraction of 90.

 So, θc=sin1μ2μ1

For total internal reflection, angle of incidence (i) at medium interface must be greater than critical angle (C).

where,

sinC=μ1μ2

Now, in given arrangement,

at point D,

sinisinr=μ2μ1 (Snell’s law) sinθsin(90C)=μ2μ1sinθcosC=μ2μ1sinθ=μ2μ1cosC=μ2μ11sin2C [from Eq. (i)] =μ2μ11μ12μ22=μ22μ121θ=sin1μ22μ121

For TIR at E,i>C

θ<sin1μ22μ121



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