Question

a concave lens has focal length of 15cm . At what distance should the object from the lens be placed so that it forms an image at 10cm from the lens ? Also , find the mignification produced by the lens.​

Answer 1

To Find :-

• The distance from the lens.

• The Magnification produced by the mirror.

Given :-

• Focal length = 15 cm

• Image distance = 10 cm

We know :-

⠀⠀⠀⠀⠀⠀⠀Lens Formula :-

$\boxed{\underline{\bf{\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}}}}$

Where :-

• f = Focal length
• v = Image distance
• u = Object distance

⠀⠀⠀⠀Magnification Fornula :-

$\boxed{\underline{\bf{m = - \dfrac{v}{u}}}}$

Where :-

• m = Magnification
• u = Image distance
• v = Object distance

Concept :-

• The values of u and f are always negative (for a Concave mirror).

• The value of v is negative for real Image and positive for virtual image.

⠀⠀⠀⠀⠀⠀⠀⠀⠀Here , the value of v will be ⠀⠀⠀⠀⠀⠀neGative as it is a real imaGe.

Solution :-

⠀⠀⠀⠀⠀⠀⠀⠀⠀Object Distance :-

Given :-

• f = 15
• v = 10

Using the lens Formula and substituting the values in it , we get :-

$:\implies \bf{\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}} \\ \\ \\ :\implies \bf{\dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f}} \\ \\ \\ :\implies \bf{\dfrac{1}{u} = \dfrac{1}{(-10)} - \dfrac{1}{(- 15)}} \\ \\ \\ :\implies \bf{\dfrac{1}{u} = \dfrac{- 3 + 2}{30}} \\ \\ \\ :\implies \bf{\dfrac{1}{u} = \dfrac{-1}{30}} \\ \\ \\ :\implies \bf{u = - 30} \\ \\ \\ \therefore \purple{\bf{u = - 30 cm}}$

Hence, the object distance (u) is - 30 cm.

⠀⠀⠀⠀⠀⠀⠀Magnification produced :-

Given :-

• v = 10
• u = - 30

Using the formula and substituting the values in it, we get :-

$:\implies \bf{m = - \dfrac{v}{u}} \\ \\ \\ :\implies \bf{m = - \dfrac{(- 10)}{(- 30)}} \\ \\ \\ :\implies \bf{m = - \dfrac{1}{3}} \\ \\ \\ :\implies \bf{m = - \dfrac{1}{3}} \\ \\ \\ \therefore \purple{\bf{m = - \dfrac{1}{3}}}$

Hence, the Magnification produced by the lens is (- ⅓).