Question

n object is placed at 58.2 cm in front of a diverging lens with a focal length of -18.1 cm. What is the magnification

Answer 1

Answer:

Explanation:

object distance u = -58.2 cm .

Focal length f = -18.1 cm .

from lens formula

$\frac{1}{v} - \frac{1}{u} =\frac{1}{f}$

Putting the values in the formula

$\frac{1}{v} + \frac{1}{58.2} =- \frac{1}{18.1}$

$\frac{1}{v} + .017=-.055$

$\frac{1}{v}$ = - .072

v = -13.80 cm

magnification

$\frac{v}{u}$

= $\frac{13.80}{58.2}$

= .237 .