Physics, asked by DopeGamer, 1 year ago

A 4.00-cm tall light bulb is placed a distance of 45.7 cm from a concave mirror having a focal length of 15.2 cm. Determine the image distance and the image size.( u= 22.8cm, h=-1.99cm)​

Answers

Answered by bhagyashreechowdhury
83

Answer: The image distance is 22.8 cm and the image size is 1.99 cm.

Explanation:

Object height, ho = 4 cm

Object distance, u = 45.7 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 mirror equation for the given concave mirror, we get

1/f = 1/v + 1/u

rearranging the equation

1/v = 1/f – 1/u

1/v = 1/15.2 – 1/45.7

1/v = 0.0657 – 0.0218 = 0.0439

v = 1/0.0439 = 22.8 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] = - [22.8 / 45.7]

hi = - [22.8 * 4] / 45.7 = - 1.99 cm

Hope this is helpful!!!!

Answered by aryanagarwal466
3

Answer:

The image distance is 22.8cm

The image size is 1.99cm

Explanation:

Item height, ho = 4 cm

Object distance, u = 45.7 cm

Focal length, f = 15.2 cm

Let the distance of the image be set to "v" cm and the length of the object and the image be identified as "ho" and "hi".

With the mirror equation for a given concave mirror we get

1 / f = 1 / v + 1 / u

repeat the equation

⇒ 1 / v = 1 / f - 1 / u

⇒ 1 / v = 1 / 15.2 - 1 / 45.7

1 / v = 0.0657 - 0.0218 = 0.0439

⇒ v = 1 / 0.0439 = 22.8 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 length (hi) and the object length (ho).

The magnification equation is expressed as follows:

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

[hoi / ho] = - [v / u]

[Hi / 4] = - [22.8 / 45.7]

⇒ hi = - [22.8 * 4] / 45.7 = - 1.99 cm

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