Question

An object 2cm in size is placed 30cm front of a concave mirror of focal length 15cm AT what di3from the mirror should screen be placed in order to obtain a sharp image find the magnification and nature of image also ​

Answer 1

Answer:

Explanation:

Given,

Object distance, u = - 30 cm

Focal length, f = - 15 cm

To Find,

Image distance, v = ?

Magnification, m =?

Formula to be used,

Mirror formula, 1/v + 1/u = 1/f

Magnification, m = h'/h = - v/u

Solution,

Putting all the values, we get

1/v + 1/u = 1/f

⇒ 1/v = 1/- 30 = 1/- 15

⇒ 1/v = 1/30 - 1/15

⇒ 1/v = 1 - 2/30

⇒ 1/v = - 1/30

⇒ v = - 30 cm

Hence, the screen should be placed at 30cm in front of the mirror.

Now, the magnification,

Magnification, m = h'/h = - v/u

⇒ m = - v/u

⇒ m = - (- 30)/- 30

⇒ m = - 1

Hence, the magnification is - 1.

Hence, the image is real and inverted.

Answer 2

Answer:

Given :-

An object 2cm in size is placed 30cm front of a concave mirror of focal length 15cm

To Find :-

Image distance, Position, Magnification

Solution :-

We know that

1/f = 1/v + 1/u

Where

f = focal length

v = image distance

u = object distance

1/-15 = 1/v + 1/-30

-1/15 = 1/v + -1/30

-1/15 = 1/v - 1/30

-1/15 + 1/30 = 1/v

-2 + 1/30 = 1/v

-1/30 = 1/v

-1v = 30

v = -30 cm

Now.

Magnification = hi/ho = -v/u

Magnification = -(-30)/-30

Magnification = 30/-30

Magnification = 1/-1

Magnification = -1

Image is real and inverted