Question

An object is placed 60cm from concave lens.the image is formed at a distance 20cm from the optical center.fi d the focal length of the lens.is the lens converging or diverging​

Answer 1

Given:

• Lens being used : concave lens

• Position of object, u = -60 cm
• Position of image, v = -20 cm

[ Position of image is negative because concave lens always form virtual and erect image ]

To find:

• Focal length of concave lens, f =?
• Nature of lens whether it is converging or diverging.

Formula required:

• Lens formula

      1 / f = 1 / v  - 1 / u

[ where f is focal length of lens, v is position of image and u is position of object ]

Calculation:

Using lens formula

→ 1 / f = 1 / v - 1 / u

→ 1 / f = 1 / (-20) - 1 / (-60)

→ 1 / f = 1 / (-20) + 1 / 60

→ 1 / f = ( - 3 + 1 ) / 60

→ 1 / f = -2 / 60

→ f = -60 / 2

→ f = -30 cm

Therefore,

• Focal length of lens is -30 cm. and
• Lens is diverging in nature.

Answer 2

$\huge\bf{\underline{\underline{\gray{GIVEN:-}}}}$

• An object is placed 60cm from a concave lens .

• The image is formed at a distance of 20cm from the optical center .

$\huge\bf{\underline{\underline{\gray{TO\:FIND:-}}}}$

1. The local length of the lens .
2. Is the lens Converging or Diverging .

$\huge\bf{\underline{\underline{\gray{SOLUTION:-}}}}$

We know that,

$\purple\bigstar\:\bf{\blue{\overbrace{\underbrace{\orange{\dfrac{1}{v}\:-\:\dfrac{1}{u}\:=\:\dfrac{1}{f}\:}}}}}$

[NOTE → The above formula is known as Lens formula or Gaussian form of lens equation .]

Where,

• $\bf\red{v}$ = Image distance

• $\bf\red{u}$ = Object distance

• $\bf\red{f}$ = Focal length

✞︎ According to the question,

• $\bf\red{v}$ = 20 cm

• $\bf\red{u}$ = 60 cm

☯︎ Here Object distance and Image distance both are taken as -ve . Because object is placed infront of the lens and image is formed also infront of the lens (Due to the lens is a concave lens) .

$\rm{{\orange{:\implies}}\:\dfrac{1}{-20}\:-\:\dfrac{1}{-60}\:=\:\dfrac{1}{f}\:}$

$\rm{{\green{:\implies}}\:-\dfrac{1}{20}\:+\:\dfrac{1}{60}\:=\:\dfrac{1}{f}\:}$

$\rm{{\orange{:\implies}}\:\dfrac{-3\:+\:1}{60}\:=\:\dfrac{1}{f}\:}$

$\rm{{\green{:\implies}}\:\dfrac{-2}{60}\:=\:\dfrac{1}{f}\:}$

$\rm{{\orange{:\implies}}\:\dfrac{1}{f}\:=\:\dfrac{-2}{60}\:}$

$\rm{{\green{:\implies}}\:f\:=\:-\dfrac{60}{2}\:}$

$\bf{{\orange{:\implies}}\:\pink{f\:=\:-30\:cm}\:}$

[NOTE → Here negative sign indicates the focal length is produced infront of the lens .]

$\huge\red\therefore$ [1] The local length of the lens is "-30cm" .

$\huge\red\therefore$ [2] It is a diverging lens (Since the focal length is negative) .