Question

please answer this. please ​

Answer 1

Answer:

Answer:

object distance,u=12cm

Given:

Radius of curvature of a concavemirror=36cm

Height of image=3×height of theobject

Focal length of the mirror,f=18cm

To find:

The position of the object

Solution:

It is given that,

height \: of \: the \: image = 3 \times height \: of \: the \: objectheightoftheimage=3×heightoftheobject

hi = 3hohi=3ho

Now,

magnification = \frac{ - v}{u} = \frac{hi}{ho} = \frac{3ho}{ho}magnification=u−v=hohi=ho3ho

m = 3m=3

\frac{ - v}{u} = 3u−v=3

v = - 3uv=−3u

By using the mirror equation,

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}f1=u1+v1

\frac{1}{18} = \frac{ - 1}{3u} + \frac{1}{u}181=3u−1+u1

\frac{1}{18} = \frac{2}{3u}181=3u2

u = 12cmu=12cm

Now,

v = - 3u = - 36cmv=−3u=−36cm

So, When the object placed between thefocus and the pole then the image willget at focal point.The image is virtualerect and magnified image.