Question

9) A converging lens has a lens power of 18 diopters. An object is placed 9.1 meters in front of the lens and an image forms at some magnification. The converging lens is then replaced with a lens with a lens power of 79 diopters and the object is moved so the image is formed in the same image position as the first lens. How many times larger is the image formed from the first lens compared to the second lens?

Answer 1

P = Power of the converging lens = 18 diopter

f = focal length of converging lens

focal length of converging lens is given as

f = 1/P

f = 1/18 = 0.056 m

d_o = object distance = 9.1 m

d_i = image distance

Using the lens equation

1/d_o + 1/d_i = 1/f

1/9.1 + 1/d_i = 1/0.056

d_i = 0.0563467 m

m = magnification = - d_i /d_o = - 0.0563467/9.1 = - 0.0062

for the new lens :

focal length of converging lens is given as

f = 1/P

f = 1/79 = 0.0127 m

d_o = object distance = ?

d_i = image distance = 0.0563467 m

Using the lens equation

1/d_o + 1/d_i = 1/f

1/d_o + 1/0.0563467 = 1/0.0127

d_o = 0.0164 m

m = magnification = - d_i /d_o = - 0.0563467/0.0164 = - 0.0062

image formed is of same size in each case.