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An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image.

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Question

An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image.

Numerical
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Solution

Giiven: Object distance (u) = −4 cm    ...(object distance is negative as per sign convention)

Focal length (f) = −12 cm    ...(Concave lens focal length is negative)

Image distance (v) = ?

By using the lens formula:

`1/v - 1/u = 1/f`

⇒ `1/v - 1/-4 = 1/-12`

⇒ `1/v = -1/12 - 1/4`

⇒ `1/v = (-3 - 1)/12`

⇒ `1/v = (-4)/12`

⇒ `v = 12/-4`

⇒ v = −3 cm 

Thus, the image is virtual and formed 3 cm away from the lens, on the left side.

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Chapter 2: Refraction of Light - Exercise 6 [Page 124]

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Lakhmir Singh Physics [English] Class 10
Chapter 2 Refraction of Light
Exercise 6 | Q 3. | Page 124
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