Question

The real image of an object is formed at a distance
of 90 cm from a concave mirror of focal length 15
cm. The magnification of the object is​

Answer 1

Explanation:

Given that,

Focal length f=−30cm

Object distance u=−90cm

Now, from the mirror formula

  f1=u1+v1

 −301+901=v1

 v1=90−2

 v=−45cm

Hence, the position and nature of image formed is -45 cm and smaller than object

Answer 2

Given:

The real image of an object is formed at a distance of 90 cm from a concave mirror

The focal length of the mirror is 15 cm

To Find:

The magnification of the object

Solution:

The magnification of the object is -5.

The magnification gives the ratio of the height of the image produced by the mirror and the height of the object.

According to the formula,

m = hi/ho = -v/u

Her v and u are the distance of the image and the object respectively from the pole of the mirror.

Using the mirror formula,

1/f = 1/v + 1/u

or 1/(-15) = 1(-90) + 1/u

or $\frac{-1}{15} + \frac{1}{90} = \frac{1}{u}$

or $\frac{1}{u} = \frac{-5}{90}$

or u = -18 cm

Substituting in the formula of magnification,

m = - (-90/-18)

= - 5

The negative sign shows that the image is real and inverted.