Question

A converging lens with a focal length of 6.50 cm forms an image of a 3.80 mm -tall real object that is to the left of the lens. The image is 2.10 cm tall and erect.

Where are the object and image located? in cm

Answer 1

Given,

f = 6.5 cm ; h = 3.8 mm ; h' = 2.1 cm

We know that

M = -(-i)/o = h'/h = 2.1/0.38 = 5.53

i = -5.53 o

from lens eqn

1/f = 1/i + 1/o

1/6.5 = 1/-5.53o + 1/o

1/6.5 = 1/o(1 + 1/-5.53) =

o = 6.5 x 0.82 = 5.33

i = 5.53 x 5.33 = -29.47 cm

Hence, o = 5.33 cm and i = -29.47 cm