Question

A concave mirror of focal length 10 cm is kept in front of an object at a distance of 50 cm in front of it if the object 1 cm high what will be the size of image

Answer 1

Given :

focal length = -10 cm

distance of the object (u) = 50cm

height of object( o) = 1cm

using the formula 1/f= 1/v +1/u

( for concave mirror );

1/10= 1/v + 1/u

1/10 = 1/v + 1/50

1/v = 1/50- 1/10

1/v = -2/25

v= -25/2

v= - 12.5cm

now using formula for magnification;

m= v/u =i/o

-12.5/ -50= i/1

12.5/50= i

i= 0.25cm

therefore the height of the image formed is o.25cm...

HOPE THIS HELPED !!!

Answer 2

Answer:

Given:-

Focal length (f) = -10 cm

Distance of the object, (u) = -50 cm

Size of the object, (O) = +1 cm

To Calculate:-

Size of the image (I) = ???

Formula to be used:-

$m=\frac{-I}{O}.......................(i)$

But to calculate I, we will have to find out m and to calculate m, we must know v.

so 1st we have to find the v.

We know that, $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

or

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

Substituting the value of f and u and we get,

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

⇒ $\frac{1}{v}=-\frac{1}{10}-(-\frac{1}{50})$

⇒ $\frac{1}{v}=-\frac{1}{10}+\frac{1}{50}$

⇒ $\frac{1}{v}=\frac{-5+1}{50}$

⇒ $\frac{1}{v}=\frac{-50}{4}$

⇒ $v=-12.5 \: cm$

Also we know that

⇒ $m=-\frac{v}{u}$

Substituting the value of v and u, we get

⇒ $m=-\frac{-12.5}{-50}=-\frac{1}{4}$

Now putting the value of m and O in Eq. (i), we get

⇒ $-\frac{1}{4}=\frac{-I}{1.0}$

⇒ $I=025 \: cm.$

Hence, the size of the image is 0.25 cm.