Question

an object 3 cm long is placed at 9 cm away from a concave mirror of focal length 18 cm find thr position and size of image​

Answer 1

${\huge{\boxed{\mathcal{\red{Answer}}}}}$

Image distance = 18cm

Size of image = 6cm

${\huge{\boxed{\mathcal{\red{Explanation}}}}}$

It is given that :-

Focal length of mirror = 18cm

Height of object= 3cm

Object distance = 9cm

Height of image = ?

Image distance = v =?

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Now according to mirror formula :

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

$f = \frac{uv}{u + v}$

In this equation :

• v = image distance
• u = object distance
• f = focal length

Now after inputting the known values in equation we get :

$18 = \frac{ - 9v}{( - 9) + v}$

$18 = \frac{ - 9v}{9 - v}$

$2 = \frac{ - v}{ 9 - v}$

$18 - 2v = - v$

$18 = - v + 2v$

$2v - v = 18$

$v = 18$

Thus the position of image from pole of mirror is 18cm.

Now we know magnification of an object is :

$\frac{h_i}{h_o}$

Magnification is also equal to :

$\frac{ - v}{u}$

Therefore :

$\frac{h_i}{h_o} = \frac{ - v}{u}$

After inputting the values in this equation we get :

$\frac{h_i}{3} = \frac{ - 18}{9}$

$\frac{h_i}{3} = - 2$

$h_i = - 6cm$

Thus the size of image is 6cm.

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BY SCIVIBHANSHU

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