If eye is kept at a depth $h$ inside the water of refractive index and viewed outside, then the diameter of circle, through which the outer objects becomes visible, be

• A

  $\frac{h}{\sqrt{{\mu }^2+1}}$
• B

  $\frac{h}{\sqrt{{\mu }^2-1}}$
• Correct answer
  C

  $\frac{2h}{\sqrt{{\mu }^2-1}}$
• D

  $\frac{h}{\sqrt{2{\mu }^2-1}}$

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