Question

an object is placed at a distance of 100cm from a concave mirror of a radius of curvature 40cm. find position,nature and size of image​

Answer 1

Given:

Distance of object = 100cm(u)

Radius of curvature = 40cm(R)

To Find:

Position,nature and size of image

Formulas Applicable:

Focal length = ½R

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

Solution:

focal length = ½ × 40cm = 20cm

Distance of image =

$\frac{1}{20} = \frac{1}{100} + \frac{1}{v}$

$= > \frac{1}{v} = \frac{1}{20} - \frac{1}{100}$

$= > \frac{1}{v} = \frac{5 - 1}{100}$

$= > \frac{1}{v} = \frac{4}{100}$

$= > \frac{1}{v} = \frac{1}{25}$

$= > v \: = 25cm$

Answer:

• Position of image is 25cm in front of mirror.

• The nature of image is real and inverted.

• The size of image is of same size of object.

Hope it helps you......✔✔✔

Answer 2

Answer:

1. The image is formed 25 cm in front of the mirror
2. The image formed is Real, Inverted and Diminished.

EXPLAINATION

Let

u = Object Distance

v = Image Distance

f = Focal Length of Mirror

R = Radius of Curvature of Mirror

m = Magnification of Image

hᵢ = Height of Image

h₀ = Height of Object

GIVEN

u = - 100 cm

R = - 40 cm

FORMULA

ᴍɪʀʀᴏʀ ғᴏʀᴍᴜʟᴀ

Object distance(u), Image distance(v) and Focal length(f) of a mirror with small aperture is related as

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

ᴍᴀɢɴɪғɪᴄᴀᴛɪᴏɴ

Magnification(m) of the image formed by a mirror is given by

$m = \frac{hi}{ho} = - \frac{v}{u}$

CALCULATION

ғᴏᴄᴀʟ ʟᴇɴɢᴛʜ

Focal length(f) and Radius of Curvature of a spherical mirror with small aperture are related as

$f = \frac{R}{2}$

$or \: f = \frac{ - 40}{2} cm$

$or \: f = - 20 \: cm$

ɪᴍᴀɢᴇ ᴅɪsᴛᴀɴᴄᴇ

Substituting values of u and f in mirror formula

$\frac{1}{v} + \frac{1}{ - 100} = \frac{1}{ - 20}$

$\frac{1}{v} = \frac{1}{100} - \frac{1}{20}$

$\frac{1}{v} = \frac{1 - 5}{100} = \frac{ - 4}{100}$

$v = - \frac{100}{4}$

$v = - 25 \: cm$

Hence, the image is formed 25 cm in front of the mirror and hence it is a real image.

ᴍᴀɢɴɪғɪᴄᴀᴛɪᴏɴ

Magnification of the image formed is

$m = \frac{hi}{ho} = - \frac{v}{u} = - \frac{ - 25}{ - 100}$

$m = - \frac{1}{4}$

⇒ Negative sign of m indicates that the image is inverted

⇒ Magnitude of m is less than 1 which indicates that the image is diminished.

Hence the image formed is Real, Inverted and Diminished.