Question

AN OBJECT OF HEIGHT 15 Cm is PLACED AT A DISTANCE 20cm from A OF CONCAVE LENS of FOCAL LENGTH 30 Cm.FIND POSITION NATURE AND Heignt of IMAGE​

Answer 1

Answer:

height of object= 15cm

u= -20cm

f= -30cm

Using Lens Formula:-

The position of image is;

$=>\frac{1}{v} -\frac{1}{u} =\frac{1}{f} \\=>\frac{1}{v} - \frac{-1}{20} =\frac{-1}{30}\\=>\frac{1}{v} + \frac{1}{20} = \frac{-1}{30}\\=>\frac{1}{v} =\frac{-1}{30} - \frac{-1}{20}\\=>\frac{1}{v} = \frac{-2-3}{60}\\=>\frac{1}{v} = \frac{-5}{60}\\=>\frac{1}{v} = \frac{-1}{12}\\=>v=-12$cm

Height of image is:-

$\frac{i}{o} =\frac{v}{u} \\=> \frac{i}{15} =\frac{-12}{-20}\\=>{i} = 9$cm

Step-by-step explanation: