Question

A real image 4/5 size of the object is formed 18 cm from a lens. Calculate the focal length of the lens.

Answer 1

$\large\underline{\bigstar \: \: {\sf Given-}}$

• Real image size is ${\sf \dfrac{4}{5}}$ of object.
• Image Distance is (v) = 18cm

$\large\underline{\bigstar \: \: {\sf To \; Find -}}$

• Focal Lenght (f)

$\large\underline{\bigstar \: \: {\sf Formula \: Used -}}$

$\implies\underline{\boxed{\sf Magnification\:(m)=\dfrac{h_i}{h_o}=\dfrac{-v}{u}}}$

${\sf h_i=Height \: of \: image }$

${\sf h_o= Height \; of \: object }$

$\large\underline{\bigstar \: \: {\sf Solution-}}$

$\implies{\sf m = \dfrac{h_i}{h_o}=\dfrac{-v}{u} }$

$\implies{\sf \dfrac{4}{5}=\dfrac{-18}{u} }$

$\implies{\sf u = \dfrac{-18\times 5}{4} }$

$\implies{\bf u = -\dfrac{45}{2}\:cm}$

We get Object Distance u = -45/2 cm

To find focal lenght -

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

$\implies{\sf \dfrac{1}{f}=\dfrac{1}{18}-\left(-\dfrac{2}{45}\right) }$

$\implies{\sf \dfrac{1}{f}=\dfrac{45+(2\times 18)}{18 \times 45} }$

$\implies{\sf \dfrac{1}{f}=\dfrac{45+36}{810}}$

$\implies{\sf f=\dfrac{810}{81}}$

$\implies{\bf f = 10\: cm}$

$\large\underline{\bigstar \: \: {\sf Answer-}}$

Focal Lenght of lens is ${\bf 10\: cm}$

Answer 2

Explanation:

Nice explanation bro awesome I also tried but I am getting wrong ans3