Question

plz help me out.......​

Answer 1

Features of image:-

NATURE - Real and inverted

MAGNIFICATION - (-1)

Magnification is equal to negative 1 this means that the image is inverted and of the same size of the object.

Now,

In case of mirrors, this case is only possible in concave mirror as in convex mirror every time diminished image is formed.

Thus,

Mirror used is a CONCAVE mirror

Now,

In concave mirror,

Object and image height remain same in the case when the object is placed at the Centre of curvature.

So,

The object is placed at the Center of curvature of the Concave mirror.

Now,

Image distance = 30cm

So,

Object distance will also be = 30cm

Now,

Center of curvature of the mirror is at 30cm. So, the focus of the mirror will be at 15cm

As we know that

$\boxed{2f = c}$

Now,

Case two :-

Object distance = 30-20 = 10cm from mirror

Focal length of mirror = 15cm

To find :- image distance

Now,

Object is placed between the Focus and pole of a concave mirror.

(As object distance <Focal length)

So,

$\boxed{u = - 10cm} \\ \\ \boxed{v = xcm} \\ \\ \boxed{f = - 15cm}$

From mirror formula we have :-

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

So

Putting values :-

$\sf{\frac{1}{v} = \frac{1}{f} - \frac{1}{u}}\\ \\ \\ \rm{\frac{1}{v} = \frac{1}{( - 15)} - \frac{1}{( - 10)}}\\ \\ \\ \sf{\frac{1}{v} = \frac{ - 1}{15} + \frac{1}{10}}\\ \\ \\ \sf{\frac{1}{v} = \frac{ - 2 + 3}{30} = \frac{1}{30}}\\ \\ \\ \boxed{v = + 30cm}$

So,

The image is formed at 30 cm behind the mirror.

The image is virtual and an erect image.

Also,

The image will be an enlarged image.

(For ray diagram, see attachment)