Question

an object 2cm long placed 40cm in front of concave mirror of focal length 15cm.find the size of the image produced

Answer 1

Answer : 1.2 cm

Given that :

Height of object = 2cm,

Distance of object = 40cm,

and focal length = 15cm

As we know that :

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ \\ = > - \frac{1}{15} = \frac{1}{v} + ( - \frac{1}{40} ) \\ \\ = > \frac{1}{v} = \frac{1}{40} - \frac{1}{15} \\ \\ = > \frac{1}{v} = \frac{3 - 8}{120} \\ \\ = > \frac{1}{v} = \frac{ - 5}{120} = \frac{ - 1}{24} \\ \\ = > v = - 24 \: cm$

As we know that :

$\frac{Height \: of \: image}{Height \: of \: object \: } = \frac{ - v}{u} \\ \\ = > \frac{Height \: of \: image}{2} = \frac{24}{ - 40} \\ \\ = > Height \: of \: image \: = \frac{2 \times 24}{ - 40} = - 1.2 \: cm$

So, Size : 1.2 cm

Nature : Real and inverted

And Position : 24 cm from concave mirror