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
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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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