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प्रश्न
A concave lens forms an erect image of `1/3`rd the size of the object which is placed at a distance 30 cm in front of the lens. Find:
- the position of image, and
- the focal length of the lens.
उत्तर
(a) m = `1/3`; u = - 30
m = `"v"/"u"`
∴ `1/3 = "v"/(- 30)`
∴ `1/cancel(3)_1 xx cancel(- 30)^(-10)` = v
∴ v = (- 10)
∴ 10 cm infront of the lens.
(b) `1/"f" = 1/"v" - 1/"u"`
`= 1/(-10) - 1/(- 30)`
`= 1/(- 10) + 1/30`
`= (- 3 + 1)/30`
`= (- 2)/30`
`= 1/"f" = - 1/ 15`
∴ f = - 15
The focal length of the given lens is 15 cm (negative).
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संबंधित प्रश्न
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