Question

The distance between an object and its image formed by a diverging lens is 49.3 cm. The focal length of the lens is -209.7 cm.
a.) Find the image distance
b.) Find the object diastance

Answer 1

Here f= - 209.7 cm and distance between image and object =49.3

So applying lens formla => 1/f = 1/v- 1/u [here v is positon of image and u is position of object]

so f= vu/ v-u

Now if postion of object is u the position of image v= u - 49.3.

so -209.7 = ( u - 49.3) u / ( u - 49.3 - u ==> u² -49.3u - 10338 =0 [quadratic equation]

By solving it we get value of u 129.27 and - 79.97

Taking value of U as -79.97 ,

B) DIstance of Oject = 79.97 cm

A) Distance of Image = 79.97- 49.3 =30.67 cm