Question

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of length 18 cm. Art what distance from the mirror should a screen be placed.so that a sharp focussed image can be obtained? Find the size and nature of the image

Answer 1

Given that, the height of the object (ho) = 7 cm

The distance of the object from the mirror (u) = -27 cm

Also, the focal length of the concave mirror (f) = - 18cm

Now,

Mirror formula:

1/f = 1/v + 1/u

→ 1/(-18) = 1/v + 1/(-27)

→ -1/18 = 1/v - 1/27

→ -1/18 + 1/27 = 1/v

→ (-3 + 2)/54 = 1/v

→ -1/54 = 1/v

Cross-multiply them

→ v = -54 cm (negative sign indicates that image is formed left side of the mirror)

Now,

m = -v/u = hi/ho

Height of image = hi and Height of object = ho

→ -(-54)/(-27) = hi/7

→ -54/27 = hi/7

→ -2 = hi/7

Cross-multiply them

→ hi = 7(-2)

→ hi = -14 cm

Therefore, the image distance from the mirror is -54 cm and height of the image is -14 cm.

Nature of the image: Real, Magnified and Inverted image.

Answer 2

Given:-

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of length 18 cm.

To find:-

• At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained?

• Find the size and nature of the image

Solution:-

Object distance (u) = -27 cm

Focal length of the concave mirror (f) = -18 cm

We know,

1/f = 1/u + 1/v

or, 1/(-18) = 1/(-27) + 1/v

or, 1/v = 1/18 + 1/27

or, $\boxed{v = - 54\ cm}$

Hence image formed at distance -54 cm . S

Now,

Magnification (m) = - image distance (v)/ object distance(u) = -v/u = Image height/ Object height = hi/ho

ho = 7 cm (given)

Therefore,

-(-54)/(-27) = (hi)/7

or, hi = (54) x 7/(-27)

or, hi = -2*7

or, hi = -14 cm

∴ So, the screen must be placed 54 cm in front of the mirror to obtain the image.

IMAGE:-

• Size - (-14 cm)

• Nature - Real and inverted

• Magnification - Enlarged