Physics, asked by poonambhaisare828, 2 months ago

A 1.2 cm long pin is placed perpendicular to the principal axis of a convex mirror of focal length 12cm. at a distance of 8cm from it.
a) find the location of the image .
b) find the height of the image.
c) Is the image erect or inverted.​

Answers

Answered by CopyThat
32

Answer:

a) 4.8 cm

b) 0.72 cm

c) erect

Explanation:

For a convex mirror,

  • f = positive
  • u = negative
  • v = positive

Where :

  • f = focal length
  • u = object distance
  • v = image distance

Here :

  • f = + 12 cm
  • u = -8 cm
  • v = ?

Using mirror formula,

  • 1/v + 1/u = 1/f
  • 1/v = 1/f - 1/u
  • 1/v = 1/12 + 1/8
  • v = 5/24
  • v = 4.8 cm

∴ a) Image is formed on the right of the mirror at a distance of 4.8 cm from it, since v is positive.

Magnification,

  • m = hₑ/hₒ = -v/u

Where :

  • hₒ = object distance
  • hₑ = image distance

Substituting :

  • m = -4.8/-8
  • hₑ = hₒ × -v/u
  • hₑ = 1.2 × 0.6
  • hₑ = 0.72 cm

∴ b) Height of the image (hₑ) is 0.72 cm.

∴ c) As hₑ is positive, image is on the same side of the principle axis as the object, so the image is erect.

Answered by peehuthakur
0

Answer:

Distance of object from mirror u=−8cm

Focal length f=12cm

We know for mirror,

f

1

=

v

1

+

u

1

12

1

=

v

1

+

−8

1

Distance of image from lens v=4.8cm

Magnification ratio m=−

u

v

=−

−8

4.8

=0.6

Height of object h=1.2cm

Height of the image h

i

=m×h=0.6×1.2=0.72cm

Explanation:

Mark me as a brainliest plz

Similar questions