Question

an object is placed at 5 cm from a concave mirror of focal length 10 cm . calculate position , nature and magnification of the image formed​

Answer 1

Given:-

• Object distance ,u = -5cm
• Focal length ,f = -10 cm

To Find:-

• Position, nature & magnification of the image

Solution:-

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀According to the Question

It is given that the object is placed 5 cm from concave mirror of focal length 10cm . we calculate the image distance using mirror formula .

• • 1/v + 1/u = 1/f

where,

• v is the Image Position
• u is the object distance
• f is the focal length

Substitute the value we get

$:\implies$ 1/v = 1/f - 1/u

$:\implies$ 1/v = 1/-10 - 1/-5

$:\implies$ 1/v = -1/10 -(-1/5)

$:\implies$ 1/v = -1/10 + 1/5

$:\implies$ 1/v = -1+2/10

$:\implies$ 1/v = 1/10

$:\implies$ v = 10cm

• Hence, the image distance is 10cm

Now, calculating the magnification of the image.

• m = -v/u

where,

• m denote magnification
• v denote image distance
• u denote object distance

Substitute the value we get

$:\implies$ m = -(10)/-5

$:\implies$ m = 10/5

$:\implies$ m = 2

• Hence, the magnification is 2 & the image formed is virtual & erect .

Answer 2

Answer:

focal length = 10 cm

object distance (u) =  -5 cm

so

1 = 1 + 1

f      u      v

$\frac{1}{10} = \frac{1}{v} +\frac{1}{-5} \\\frac{1}{10} -\frac{1}{-5} =\frac{1}{v} \\\frac{1+2}{10} =\frac{1}{v} \\\frac{3}{10}=\frac{1}{v} \\\frac{10}{3} = v= 3.33$

m = -v/u = -(10/3)/-5 = 10/15 = 2/3

so the position of the image is 3.33  cm behind the mirror

the image will be enlarge with a magnification of 2/3 .

hope it help