Question

And object of size 3 cm is placed at a distance of 15 cm from a convex lens of focal length 10 cm calculate the distance and size of the image so formed. Also write nature of the image

Answer 1

given, u=-15cm,

f=-10cm( it is a convex lens)

using lens formula,

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ = > - \frac{1}{10} = \frac{1}{v} - ( - \frac{1}{15} ) \\ = > - \frac{1}{10} = \frac{1}{v} + \frac{1}{15} \\ = > \frac{1}{v} = - \frac{1}{10} - \frac{1}{15} \\ = > \frac{1}{v} = \frac{ - 3 - 2}{30} \\ = > \frac{1}{v} = \frac{ - 5}{30} \\ = > v = - \frac{30}{5} = - 6$

so image will formed at a distance of -6 cm in front of the lens.

$\bold{Nature\:of\:image}$ - the image is real and inverted.

we have,

size of object,$h_1$= 3cm

we know that ,

$\frac{v}{u} = \frac{h_{2}}{ h_{1}}$

so,putting values of v, u ,$h_2$in above formula,

$\frac{ - 6}{ - 15} = \frac{ h_{2} }{3} \\ = > \frac{6}{15} = \frac{ h_{2} }{3} \\ = > h_{2} = \frac{18}{15} = 1.2 \: cm$

so,size of the image is 1.2 cm.

hope this helps.