Question

A lens which has focal length of 4cm and refractive index1.4 is immersed ina liquid of refractive index1.6 ,then focal length

Answer 1

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

• Focal Lenght of lens ${\sf f_1}$= 4cm and refractive index $\mu_1$ = 1.4
• Lens is immersed in a liquid of refractive index $\mu_2$ = 1.6

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

• Focal Lenght after the lens is immersed in water ${\sf (f_2)}$

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

$\color{violet}\bullet\underline{\boxed{\sf f=(\mu-1)×\left[\dfrac{1}{R_1}-\dfrac{1}{R_2}\right]}}$

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

$\implies{\sf \dfrac{f_1}{f_2}=\dfrac{\mu_1-1}{\mu_2-1} }$

$\implies{\sf \dfrac{4}{f_2}=\dfrac{1.4-1}{1.6-1}}$

$\implies{\sf \dfrac{4}{f_2}=\dfrac{0.4}{0.6}}$

$\implies{\sf \dfrac{4}{f_2}=\dfrac{2}{3} }$

$\implies{\sf f_2=\dfrac{4×3}{2}}$

$\implies{\sf f_2=\dfrac{12}{2}}$

$\color{red}\implies{\sf f_2=6\:cm }$

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

Focal Lenght after the lens is immersed in liquid of refractive index 1.6 will be $\color{red}{\sf 6\:cm}$