Question

an object is placed at a distance of 12 cm from a concave lens of focal length 26cm. where do you get the image. state the nature and size of the image formed

Answer 1

The type of lens used is concave.

→ 1/f = 1/v - 1/u

• u = - 12
• f = - 26

→ 1/(- 26) = 1/v - 1/(- 12)

→ - 1/26 = 1/v + 1/12

→ - 1/26 - 1/12 = 1/v

→ (- 6 - 13)/156 = 1/v

→ - 156/19 = v

We will get image in negative side, side where object is placed.

→ m = v/u = - 156/19 × 1/( - 12)

→ m = 13/9

Nature: The image formed is virtual.

Size: The image is magnified.

Answer 2

» An object is placed at a distance of 12 cm from a concave lens of focal length 26cm.

Here..

Object distance from lens (u) = - 12 cm

Focal length (f) = - 26 cm

__________ [ GIVEN ]

• We have to find the image distance from lens, magnification, nature and size of the image formed.

_____________________________

We know that..

$\dfrac{1}{f}$ = $\dfrac{1}{v}\:-\:\dfrac{1}{u}$

[ LENS FORMULA ]

=> $\dfrac{1}{-26}$ = $\dfrac{1}{v}\:-\:\dfrac{1}{-12}$

=> $\dfrac{-1}{26}$ = $\dfrac{1}{v}\:-\:\dfrac{(-1)}{12}$

=> $\dfrac{-1}{26}$ = $\dfrac{1}{v}\:+\:\dfrac{1}{12}$

=> $\dfrac{1}{v}$ = $\dfrac{-1}{26}$ - $\dfrac{1}{12}$

=> $\dfrac{1}{v}$ = $\dfrac{-6\:-\:13}{156}$

=> $\dfrac{1}{v}$ = $\dfrac{-19}{156}$

=> v = $\dfrac{-156}{19}$ cm

So.. image is placed in the same side as that of the object. (in negative side).

_____________________________

Now..

m = $\dfrac{-v}{u}$

=> $\dfrac{ \frac{ - 156}{19} }{ - 12}$

=> $\dfrac{-156}{19}$ × $\dfrac{-1}{12}$

=> $\dfrac{13}{19}$

=> 0.68

The image formed is virtual and large in size.

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