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?

Asked by  | 1st Feb, 2014, 04:19: PM

Expert Answer:

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.

Answered by  | 3rd Apr, 2014, 12:45: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.