Question

2. An object 3cm high is placed at a distance of 10cm in front of a concave mirror
of focal length 20cm. Find the position, nature and size of the image formed.​

Answer 1

Explanation:

$height \: of \: the \: object = 3cm \\ \\ object \: of\: distance \: from \: the \: mirror = - 10cm \\ \\ focal \: length = - 20 \\ the \: image \: height = hi \\ \\ the \: image \: distance = v \\ \\ \\ \\ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ \\ \frac{1}{ -20} = \frac{1}{v } + \frac{1}{ - 10} \\ \\ - \frac{1}{20} = \frac{1}{v} - \frac{1}{10} = \frac{1}{v } = \frac{1}{ - 20} + \frac{1}{10 } \\ \frac{1}{v} = \frac{( - 1 + 2)}{20} = \frac{1}{20} \\ v = + 20 \\ magnification \: of \: the \: mirror = \frac{hi}{ho} = \frac{ - v}{u} \\ \\ \\ \frac{hi}{3} = \frac{ - 20}{ - 10} \\ \\ \frac{hi}{3} = \frac{2}{1 } \\ hi = 6cm$

if v is + be than the image is virtual and erect , upright

Answer 2

Explanation:

Given : h1 = 3 cm, u = - 10 cm, f = - 20 cm

We know that

1/v + 1/u = 1/f

⇒ 1/v = 1/f – 1/u = 1/(-20) – 1/(-10)

= - (1/20) + 1/10 = (-1 + 2)/20 = 1/20

∴ v = 20 cm

The image is formed at a distance of 20 cm behind the mirror.

And

m = - (v/u) = h2/h1

⇒ - (20)/(-10) = h2/3

⇒ h2 = 6 cm

Image is 6 cm in size, virtual and erect.