Question

An object of height 4.25 mm is placed at a distance of 10 cm from a convex lens of power +5D. Find (i) the focal length of the lens, and (ii) the size of the image.

Answer 1

Object height, h = 4.25 mm = 0.425 cm (1 cm = 10 mm)
Object distance, u =-10 cm
Power, P = +5 D
Focal length, f = ?
Image distance, v = ?
Image height, h' = ?  

Power, P ${\displaystyle =\frac{1}{f}}$ 

${\displaystyle f=\frac{1}{p}=\frac{1}{5}=0.2m=20cm}$ 

Using the lens formula, we get: 

${\displaystyle \frac{1}{v}-\frac{1}{u}=\frac{1}{f}}$ 

${\displaystyle \frac{1}{v}-\frac{1}{-10}=\frac{1}{20}}$ 

${\displaystyle \frac{1}{v}+\frac{1}{10}=\frac{1}{20}}$ 

${\displaystyle \frac{1}{v}=\frac{1}{20}-\frac{1}{10}}$ 

${\displaystyle \frac{1}{v}=-\frac{1}{20}}$ 

∴ v=-20cm 

Now, magnification, m = ${\displaystyle \frac{h\prime }{h}=\frac{v}{u}}$ 

Substituting the values in the above equation, we get: 

${\displaystyle \frac{h\prime }{0.425}=\frac{-20}{-10}}$ 

${\displaystyle h\prime =2X0.425=0.85cm=8.5mm}$ 

Thus, the image is 8.5 mm long; it is also erect and virtual.