An object is kept at a distance of 45cm from a screen. where should a converging lens of focal length 10cm be kept in order to obtain a sharp image on the screen?

Focal length of the convex lens = + 10cm

Let Object distance = u 

Let image distance = v 

Given, u + v = 45 cm

or, v = 45 - u

 

From lens formula,

 

1/f = 1/v - 1/u

or, 1/10= (u - v)/ uv

or, 10(u - v) = uv

or, 10 [ u - ( 45-u)] = u (45 - u)

or, 10( 2u - 45) = 45 u - u²

or, 20u - 450 = 45 u -u²

or, u² - 45 u +20u -450 = 0

or, u² - 25 u - 450 = 0

 

Solving the above equation, we get u = 71.3 cm or -6.3cm

Since the object distance is negative for concave lens, u = -6.3 cm

Therefore, v = 45 - u = [45 - (6.3)] cm

or, v = 38.7 cm

 

Thus, the converging lens should be kept at a distance of 6.3 cm to the right of the object, so as to obtain a sharp image on screen.