Q142 / 160 · AP EAMCET 2023 (19 May Shift 1) · Physics

With respect to air, the critical angle in a medium for red light of wave length $\lambda_1$ is $\theta$. Other facts remaining same, critical angle for yellow light of wave length $\lambda_2$ will be

  • A
    $\theta$
  • B
    more than $\theta$
  • Correct answer
    C
    less than $\theta$
  • D
    $\frac{\theta \lambda_1}{\lambda_2}$

Solution

Critical angle $=\sin ^{-1}\left(\frac{1}{\mathrm{n}}\right)=\sin ^{-1}\left(\frac{\mathrm{v}}{\mathrm{c}}\right)$ Critical angle $=\sin ^{-1}\left(\frac{\mathrm{f} \lambda}{\mathrm{c}}\right) \quad\left(\because \mathrm{v}=\frac{\mathrm{c}}{\mathrm{n}}\right)$ for red light, $\theta=\sin ^{-1}\left(\frac{f \lambda_1}{c}\right)(\because v=f \lambda)$ for yellow light $\theta^{\prime}=\sin ^{-1}\left(\frac{\mathrm{f} \lambda_2}{\mathrm{c}}\right)$ Since $\lambda_1>\lambda_2$ then $\theta^{\prime} < \theta$

Hence, the correct answer is option: 'C'.

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