Question

An object is located a distance 3f/2 from a thin converging lens of focal length f as shown in the diagram below.

a. Calculate the position of the image.

b. Trace two of the principal rays to verify the position of the image.

c. Suppose the object remains fixed and the lens is removed. Another converging lens of focal length f₂ is placed in exactly the same position as the first lens. A new real image larger than the first is now formed. Must the focal length of the second lens be greater or less than f ? Justify your answer

Answer 1

(a) 1/f = 1/d_o + 1/d_i … 1/f = 1/ (3f/2) + 1/d_i … d_i = 3f

(b)

(c) The new lens makes a larger magnification. M = – d_i / d_o → with the same do, di must have increased. Then based on 1/f = 1/d_o + 1/d_i, to get a larger di with the same do the focal point must be larger. You can test this with sample numbers if you don’t see it at first.