Question

Ab object is placed at a distance 'U' from a converging lens produced a virtual image 33cm from the object. If the magnification is 4, calculate the:

a)value of U

b)Focal length of the lens

c)Power of the lena

Answer 1

a) From the Magnification equation, we have:

$$M=\dfrac{-v}{u},$$ $$u=\dfrac{-v}{M}=\dfrac{-(-33\ cm)}{4}=8.25\ cm.$$

b) From the thin lens equation, we get:

$$\dfrac{1}{u}+\dfrac{1}{-v}=\dfrac{1}{f},$$ $$f=\dfrac{1}{\dfrac{1}{u}-\dfrac{1}{v}},$$ $$f=\dfrac{1}{\dfrac{1}{8.25\ cm}-\dfrac{1}{33\ cm}}=11\ cm.$$

c) By the definition of the lens power, we get:

$$P=\dfrac{1}{f}=\dfrac{1}{11\ cm}=0.1\ D.$$