A concave lens forms an erect image of 1/3rd size of the object which is placed at a distance 30 cm in front of the lens. Find - the position of image the focal length of the lens. - Physics

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Numerical

A concave lens forms an erect image of 1/3rd size of the object which is placed at a distance 30 cm in front of the lens. Find -

  1.  the position of image, and
  2. the focal length of the lens.
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Solution

(a) m = `1/3`; μ = - 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).

Concept: Concave Lens
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Chapter 5: Refraction through a Lens - Exercise 5 (C) 2 [Page 127]

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Selina Concise Physics Class 10 ICSE
Chapter 5 Refraction through a Lens
Exercise 5 (C) 2 | Q 10 | Page 127

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