Question

A typical human lens has an index of refraction of 1.430 . The lens has a double convex shape, but its curvature can be varied by the ciliary muscles acting around its rim. At minimum power, the radius of the front of the lens is 10.0 mm, whereas that of the back is 6.00 mm. At maximum power, the radii are 6.50 mm and 5.50 mm, respectively. If the lens were in air, what would be the maximum power and associated focal length of the lens

Answer 1

Answer:

Maximum Power = 144.3 D

The associated focal length of the lens = $6.92*10^{-3} m$

Explanation:

According to the Lens maker's Formula:

$\frac{1}{f} = (n-1) (\frac{1}{R_1}-\frac{1}{R_2} )$

where;

$n_1$ = the refractive index of the medium

$R_1$ and $R_2$ = radius of curvature on each surface

For a convex lens, The radius of curvature in the front surface will be positive and that of the second surface will be negative . Therefore;

$\frac{1}{f} = (n-1) (\frac{1}{R_1}-\frac{1}{-R_2} ) \\ \\ \frac{1}{f} = (n-1) (\frac{1}{R_1}+\frac{1}{R_2} )$

At maximum power

$\frac{1}{f} = (1.430-1) (\frac{1}{6.50 \ mm}-\frac{1}{5.50 \ mm} )$

= $0.144 \ mm^{-1}$

This Implies

$f = 6.92 mm\\f = 6.92*10^{-3} \ m$

Therefore; the power is given by the formula:

$P_{max} = \frac{1}{f}$

$P_{max}= \frac{1}{6.92*10^{-3}}$

= 144.3 D