Radius of curvature of human eye is 0.78 cm For an object at infinity, image is formed at 3 cm behind the refracting surface. The refractive index of eye is

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6.2

1.35

answer is A.

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Detailed Solution

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$
Putting v = 3 cm
$\begin{array}{l} μ_{1} = 1 \\ u = \infty \end{array}$
u = 0.78cm
$\begin{array}{l} \frac{μ_{2}}{3} - \frac{1}{\infty} = \frac{μ_{2} - 1}{0 .78} \\ ∴ μ_{2} = 1 .35 \end{array}$

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