Physics, asked by navinpant123, 8 months ago

An object of 1cm height produces a real image 1.5cm high when placed at a distance of 15 cm from a concave mirror . Calculate the position of image and the focal length of the mirror​

Answers

Answered by Cosmique
13

Given :

  • height of object , h₁ = 1 cm
  • image produced by concave mirror is real and inverted
  • height of image , h₂ = -1.5 cm   [ negative height of image, because image formed is real ]
  • position of image, u = -15 cm

To find :

  • position of image, v = ?
  • focal length of image, f = ?

Formulae required :

  • Formula for magnification produced by mirror

    m = - v / u = h₂ / h₁

  • Mirror formula

    1/f = 1/v + 1/u

[ where m is magnification. v is position of image, u is position of object, f is focal length of mirror, h₁ is height of object , h₂ is height of image ]

Solution :

Using formula for magnification of mirror

→ m = - v / u = h₂ / h₁

→ - v / u = h₂ / h₁

→ - v / ( -15 ) = (-1.5) / (1)

→ v / (15) = -1.5

v = -22.5 cm

Using mirror formula

→ 1/f = 1/v + 1/u

→ 1/f = 1/(-22.5) + 1/(-15)

→ 1/f = 10/(-225) + 1/(-15)

1/f = (-10 - 15) / (225)

→ 1/f = (-25) / (225)

→ f = 225 / -25

f = -9 cm

therefore,

  • position of image is 22.5 cm infront of the concave mirror.
  • focal length of concave mirror is -9 cm . (focal length of concave mirror is always negative) .
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