Question

an object is placed at distance of 20cm in front of convex mirror of radius of curvature 30cm. find the position and nature of the image

Answer 1

Answer:

u = - 20 cm

r = 30

we know that

f= r/2

f = 15

by mirror formula

1/ f = 1/ v + 1/ u

1/ v = 1/ f - 1/ u

1/ v = 1/ 15 - 1/ -20

1/ v = 7/ 60

v = 60/7

v = 8.5 cm.

terefoer the image formed is virtual and errect

thanks I hope it will help you :)

Answer 2

Given,

Distance of object = 20 cm.

Radius of curvature = 30 cm.

To Find,

Position and nature of the image.

Solution,

We can solve this mathematical problem using mathematical function.

Let distance of object be 'u'

Focal lenght ( f) = $\frac{Radius of curvature}{2}$

⇒ $\frac{30}{2}$ ⇒ 15cm.

We know apply Mirror formula,

⇒  $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

⇒ $\frac{1}{15} = \frac{1}{v} + \frac{1}{-20}$

⇒ $\frac{1}{v} = \frac{1}{15} - \frac{1}{20}$

⇒ $\frac{1}{v} = \frac{4-3}{60}$

⇒ $\frac{1}{v} = \frac{1}{60}$

⇒ v = 60cm.

We know that,

⇒ $\frac{h'}{h} = \frac{-v}{u}$

⇒ $h' = \frac{-v}{u} * h$

⇒ $h' = \frac{-60}{-20} * 15$

Hence, h' = 15 cm.

Therefore, the postion of image(v) = 60 cm.

And, the nature of image is virtual and erect.