# 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

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.

#### 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]

#### APPEARS IN

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