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Wave Optics question

2004 · Shift 0 · Q126
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Wave Optics question

2004 · Shift 0 · Q126

JEE MainPhysicsWave OpticsMCQ+4 / −1
The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index nnn) is :
  1. A
    tan⁡−1(1/n){\tan ^{ - 1}}\left( {1/n} \right)tan−1(1/n)
  2. B
    sin⁡−1(1/n){\sin ^{ - 1}}\left( {1/n} \right)sin−1(1/n)
  3. C
    sin⁡−1(n){\sin ^{ - 1}}\left( n \right)sin−1(n)
  4. D
    tan⁡−1(n){\tan ^{ - 1}}\left( n \right)tan−1(n)
View written solutionFree

Correct answer: D

  1. For light going from air to glass, the angle at which the reflected light is completely plane polarized is called the Brewster angle iBi_BiB​.

  2. Brewster's law states: tan⁡iB=n2n1\tan i_B = \frac{n_2}{n_1}taniB​=n1​n2​​ where n1n_1n1​ is the refractive index of the first medium and n2n_2n2​ is that of the second medium.

  3. Here, light travels from air to glass:

  • n1=1n_1 = 1n1​=1
  • n2=nn_2 = nn2​=n

So, tan⁡iB=n1=n\tan i_B = \frac{n}{1} = ntaniB​=1n​=n

  1. Therefore, iB=tan⁡−1(n)i_B = \tan^{-1}(n)iB​=tan−1(n)

  2. Now compare with the options:

  • A: tan⁡−1(1/n)\tan^{-1}(1/n)tan−1(1/n) — incorrect
  • B: sin⁡−1(1/n)\sin^{-1}(1/n)sin−1(1/n) — incorrect
  • C: sin⁡−1(n)\sin^{-1}(n)sin−1(n) — invalid for n>1n>1n>1
  • D: tan⁡−1(n)\tan^{-1}(n)tan−1(n) — correct

Hence, the correct answer is: tan⁡−1(n)\boxed{\tan^{-1}(n)}tan−1(n)​

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