Question

A bright object 50 mm high stand in the axis of concave mirror of focal length 100 nn and at the distance of 300 mm from concave mirror .How big will the image be?

Answer 1

Explanation:

1/v + 1/u=1/f

1/v+1/-300= 1/-100

1/v= 1/-100-1/300

1/v= -2/300

v= -150.

m=h'/h= -v/u

h'= -hv/u

h'= -50*-150/300

h'=25

hope it helps u...

Answer 2

Ho= 50mm=5cm

f=100 mm=-10 cm

u=300 mm=-30cm

$\frac{1}{f } = \frac{1}{v} + \frac{1}{u} \\ \frac{ - 1}{10} + \frac{1}{30} = \frac{1}{v} \\ \frac{ - 20}{300} = \frac{1}{v} \\ v = - 15cm$

$m = \frac{ - v}{u} = \frac{hi}{ho} \\ hi = \frac{ - ( - 15) \times 5}{ - 30} = - 2.5cm$

I hope this helps ( ╹▽╹ )