Question

A convex lens of focal length 10 cm produces a real image that is 3 times the size of an object. What will be the distance of the object from the lens?​

Answer 1

Given :-

• focal length of convex lens = + 10 cm
• image formed is real
• magnification of image , m = -3

(focal length of convex lens is always positive )

( magnification is negative since image formed is real )

To find :-

• distance of object from lens

Formulae Required :-

Formula for magnification produced by lens

$\red{\bigstar}\boxed{\sf{m=\dfrac{v}{u}}}$

(where m is the magnification , u is the position of object , v is the position of image.)

Lens Formula

$\red{\bigstar}\boxed{\sf{\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}}}$

(where f is the focal length of lens , v is the position of image , u is the position of object)

Solution :-

⇢  position of object = u

⇢  position of image = v

⇢  magnification produced = m = - 3

⇢  focal length of lens = f = + 10 cm

▶Using the formula for calculating magnification of lens

$\implies\sf{m=\dfrac{v}{u}}\\\\\implies \sf{-3=\dfrac{v}{u}}\\\\\implies\boxed{\sf {v=-3u\;\;\;\;....equation(1)}}$

▶Using the lens Formula

$\implies\sf{\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}}\\\\\implies\sf{\dfrac{1}{10}=\dfrac{1}{v}-\dfrac{1}{u}}$

using equation (1)

$\implies\sf{\dfrac{1}{10}=\dfrac{1}{(-3u)}-\dfrac{1}{u}}\\\\\implies\sf{\dfrac{1}{10}=\dfrac{1-(-3)}{-3u}}\\\\\implies\sf{-3u=40}\\\\\implies\boxed{\boxed{\sf{u=-\;\dfrac{40}{3}\;\;cm=-13.34\;\;cm}}}$

Hence,

Distance of object from the Lens is 13.34 cm in front of the lens .

Answer 2

Given,

Focal length (f) = 10 cm

Magnification (m) = -3

Here, the magnification is taken negative since the image formed is real.

To Find,

The distance of an object from the lens

Solution,

Magnification produced by the lens is given by,

       $m=\frac{v}{u}$

Here, 'v' is the position of the image and

'u' is the position of object.

Substituting the given value, we get

       $-3=\frac{v}{u}$

       $v=-3u$          ⇒ Equation (1)

Using the lens formula,

       $\frac{1}{f} = \frac{1}{v}-\frac{1}{u}$

       $\frac{1}{10} = \frac{1}{v}-\frac{1}{u}$

Substituting the equation (1) in above equation,

       $\frac{1}{10} = \frac{1}{-3u}-\frac{1}{u}$

       $\frac{1}{10} = \frac{1-(-3)}{-3u}$

     $-3u = 40$

       $u =-\frac{40}{3}$

       u = -13.34 cm

Hence,

The distance of the object from the Lens is 13.34 cm in front of the lens.