Question

3. An object 4 cm high is placed at a distance of 10cm from a convex lens of focal length 20cm. Find the position nature and size of an image​

Answer 1

Answer:P:-behind the object

S:-enlarged

N:-virtual & erect

Explanation:

$\frac{1}{F} =\frac{1}{V} -\frac{1}{U}\\\\\frac{1}{F} +\frac{1}{U} =\frac{1}{V}\\\frac{1}{20} +\frac{1}{-10} =\frac{1}{U}\\\frac{-1}{20} =\frac{1}{U} \\U= -20$