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Geometrical Optics question

2020 · 8 Jan · Shift 1 · Q52
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Geometrical Optics question

2020 · 8 Jan · Shift 1 · Q52

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability 43{4 \over 3}34​ for this wavelength, will be :
  1. A
    45°
  2. B
    15°
  3. C
    30°
  4. D
    60°
View written solutionFree

Correct answer: C

  1. Refractive index of the medium

For a medium,

n=μrϵrn = \sqrt{\mu_r \epsilon_r}n=μr​ϵr​​

Given:

ϵr=3,μr=43\epsilon_r = 3, \qquad \mu_r = \frac{4}{3}ϵr​=3,μr​=34​

So,

n=3⋅43=4=2n = \sqrt{3 \cdot \frac{4}{3}} = \sqrt{4} = 2n=3⋅34​​=4​=2

  1. Critical angle formula

For light going from denser medium (n1=2)(n_1=2)(n1​=2) to air (n2=1)(n_2=1)(n2​=1), the critical angle CCC satisfies:

sin⁡C=n2n1=12\sin C = \frac{n_2}{n_1} = \frac{1}{2}sinC=n1​n2​​=21​

Thus,

C=sin⁡−1(12)=30∘C = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circC=sin−1(21​)=30∘

  1. Match with options
  • A: 45∘45^\circ45∘ ❌
  • B: 15∘15^\circ15∘ ❌
  • C: 30∘30^\circ30∘ ✅
  • D: 60∘60^\circ60∘ ❌

Hence, the correct option is C.

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