Question

An object 4.0 cm tall is placed in front of a converging lens whose focal length 22 cm. Where is the image formed, and what are its characteristics, if the object distance is

(a) 15 cm

(b) 36 cm?

Answer #1

a) Using lens equation

1/f =1/q +1/p

1/q =1/f -1/p

Here p= 15 cm , f=22 cm

1/q = 1/22 -1/15

1/q = (15-22)/15×22

1/q =-7/15×22

q=-15×22/7

q= -47.143 cm (image distance)

(- show that the image is virtual )

Magnification

M=-q/p =-(-47.143)/15 = 3.143

Since M is positive so the image is upright . And M>1 so the image is larger than the object.

h'=Mh =3.143×4=12.572 cm ( image size)

b) similarly

1/q = 1/22 -1/36

1/q = (36-22)/(22×36)

q= 22×36/14

q = 56.57 cm

Since q is positive so the image is real.

M=-q/p =-56.57/36

M=-1.57

Since M is negative so the image is inverted .

h' =Mh

h' =1.57×4

h' = 6.28 cm ( image size.)

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