Question

A real object is imaged by a thin lens (F=48mm). The image is located 300 mm from

the object in air. Determine the possible locations of the lens. Determine the

magnification and whether the image is upright or inverted and magnified or minified.

What is the longest focal length lens that would produce 300 mm separation between

the object and the image?

Answer 1

For object distance 'u', image distance 'v' and focal length 'f' such that distance

(this thin lens equation is as per cartesian sign conventions)

for a given value of u, from the constraint of distance between object and image

thus the thin lens equation becomes

i.e there are two cases: (i) when u = 240 mm before lens - thus v = 60 mm after lens. the image is inverted and minified. with magnification m = v/u = 60/240 = 1/4 = 0.25

and (ii) when u = 60 mm before lens implies v = 240 mm after lens. The image is inverted and magnified with the value of linear magnification m = 240/60 = 4

For the longest focal length that can give a separation of 300 mm between object and image, the formula is that the distance should be equal to 4f. with u = v = 2f.

i.e. u = v = 2f. given u + v = 300 mm. thus, u = 150 mm

i.e. 2f = 150 mm,   f = 75 mm

the derivation can be seen below if required

the solution of a quadratic equation is

for the solution to have real values (not imaginary)

for the case of minimum value of focal length f