Question

An object 1.55 cm high is held 2.95 cm from a person’s cornea, and its reflected image is measured to be 0.175 cm high.

a) What is the magnification?

b) What is the image distance in cm?

c) Find the radius of curvature in cm of the convex mirror formed by the cornea.

Answer 1

Given that,

Height of the object, $\small H_o$ = 1.55 cm

Height of the image, $\small H_i$ = 0.175 cm

Object distance from cornea, u = 2.95 cm

a) Magnification of the convex mirror (m )

Magnification of the mirror is defined as the ratio of image height to object height

$\small m=\frac {H_i} {H_o}$

$\small m=\frac {0.175(cm)} {1.55(cm)} =\bold {0.1129}$

Therefore Magnification, m = 0.1129

b) Images distance from the mirror ( v.)

Magnification of the convex mirror is also expression as the negative ratio of image distance to object distance

$\small m=\frac {-v} {u}$

$\small 0.1129=\frac {-v} {2.95(cm)}$

V= -0.1129 * 2.95

v = -0.333

c) Radius of curvature of convex mirror (R)

We know the mirror equation,

$\small \frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

Where f is the Focal length of the mirror

By substituting the values of v and u

$\small \frac{1}{f}=\frac{1}{2.95(cm)}-\frac{1}{0.333(cm)}$

$\small \frac{1}{f}=-2.66$

Therefore, Focal length, f = -0.376 cm

Radius of curvature is twice the Focal length of mirror. So radius of curvature of convex mirror,

R = 2 * f

R = 2 * (-0.376)

R = -0.752 cm

Radius of curvature, R = -0.752 cm