Question

The human eye can be regarded as a single spherical refractive surface having curvature of cornea $7.8\text{mm}$. If a parallel beam of light comes to focus at $3.075\text{cm}$ behind the refractive surface, the refractive index of the eye is

• A. $1.34$
• B. $1.72$
• C. $1.5$
• D. $1.61$

Answer 1

The correct option is A $1.34$


Sign convention: Direction of incident ray taken as +ve.

Here object distance, $u=-\infty$
Radius of curvature of spherical surface (cornea),
$R=+7.8\text{mm}=+0.78\text{cm}$
Image distance, $v=+3.075\text{cm}$

Hence,
$\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}$

$\Rightarrow \frac{\mu }{+3.075}-\frac{1}{-\infty }=\frac{\mu -1}{+0.78}$

or, $\frac{\mu }{3.075}-0=\frac{\mu -1}{0.78}$

or, $\frac{\mu -1}{\mu }=\frac{0.78}{3.075}$

or, $1-\frac{1}{\mu }\simeq 0.25$

or, $\frac{1}{\mu }=1-0.25\approx 0.75$

$\Rightarrow \mu =\frac{1}{0.75}=\frac{4}{3}=1.333$

$\therefore \mu \approx 1.33$

Why this question? Tip: Cornea has been given as spherical surface and image is formed where light rays are focused (converged) after refraction.