Question

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

Answer 1

$\huge\mathfrak\green{Answer:}$

Given:

We have been given that the object size(h) = +7.0cm, object distance(u) = -27cm, focal length(f) = -18cm.

To Find:

We need to find the image distance(v) and image size(h'). We also need to find the nature of the image.

Solution:

As it is given that object distance(u) = -27cm and the focal length(f) = -18cm, therefore we can find the image distance by the mirror formula.

1/v + 1/u = 1/f

=> 1/v + 1/-27 = 1/-18

=> 1/v = 1/-18 - 1/-27

=> 1/v = (-3 + 2)/54

=> 1/v = -1/54

=> v = -54cm

Magnification = h/h' = -v/u

= -(-54)/(-27)

= 54/(-27)

= -2

Now, we need to find the Image size, we have

h' = -vh/u

=> {-(-54) × (7)}/(-27)

= (54 × 7)/(-27)

= 378/(-27)

= -14cm

Therefore, the image is real, inverted and enlarged in size.

Answer 2

Answer:

Explanation:

Given :-

Object distance, u = - 27 cm

Focal length, f = - 18 cm

To Find :-

Image formed, v = ??

Size and the nature of the image.

Formula to be used :-

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

Magnification, m = - v/u = h₂/h₁

Solution :-

Putting all the values, we get

1/f = 1/u + 1/v

⇒  1/(-18) = 1/(-27) + 1/v

⇒  1/v = 1/18 + 1/27

⇒ 1/v = - 1/54

⇒  v = - 54 cm

Now, Magnification,

Magnification, m = - v/u = h₂/h₁

⇒ h₂ = - v × h₁/u

⇒ h₂ = - 54 × 7/- 27

⇒ h₂ = - 14 cm

Hence, image is real, inverted and enlarged in size.