Question

A 4cm tall light bulb is placed at a distance of 8.3 from a concave mirror having a focal length 15.2cm determine the image distance and the image sixe​

Answer 1

Given

h=4cm

u= -8.3cm

f= -15.2cm(for concave mirror)

By mirror formula

1/v + 1/u = 1/f

1/v = 1/f - 1/u

1/v = 1/(-15.2) - 1/(-8.3)

1/v = -1/15.2 + 1/8.3

= -10/152 + 10/83

= (-830+1520)/12616

= 690/12616

v=12616/690=18.28cm

Answer 2

Answer:

Hi,

Answer:

The image distance v = -18.3 cm and the image size hi = 8.8 cm.

Explanation:

Object height, ho = 4 cm

Object distance, u = 8.3 cm

Focal  length, f = 15.2 cm

Let the image distance be denoted as “v” cm and the height of object and image be denoted as “ho” & “hi” respectively.

Using the lens formula for the given concave mirror, we get

1/f = 1/v + 1/u

⇒ 1/v = 1/f – 1/u

⇒ 1/v = 1/15.2 – 1/8.3

⇒ 1/v = 0.0657 – 0.1204 = - 0.0547

⇒ v = 1/0.0547

⇒ v = - 18.28 cm ≈ 18.3 cm

The magnification equation is defined as the ratio of the image distance (v) and the object distance (u) to the ratio of the image height (hi) and the object height (ho). The magnification equation is stated as follows:

M = [hi / ho ] = - [v / u]

∴ [hi / ho ] = - [v / u]

⇒ [hi / 4] = - [(-18.3) / 8.3]

⇒ hi = [18.3 * 4] / 8.3 = 8.8 cm

Hope this helpful!!!!!!