Question

A convex lens have focal length of 20 cm. the material of lens has refractive index of $1.5$ is immersed in water of refractive index 4/3. find the change in focal length of the lens.

Answer 1

$\frac{1}{fa}=(ang-1)(\frac{1}{R_1}-\frac{1}{R_2})---\left(1\right)$

give that $ang=1.5=$refrentiue indent of gases w.r.t air

$fa=$ focal length in air

$\Rightarrow ft=$ focal length in liquid will be given

$\frac{1}{ft}=(lng-1)(\frac{1}{R_1}-\frac{1}{R_2})---\left(2\right)$

dividing $\left(1\right)$ & $\left(2\right)$ we get,

$\frac{ft}{fa}=\frac{(ang-1)}{(lng-1)}=\frac{(1.5-1)}{\left(\frac{ang}{anl}\right)}=\frac{0.5}{(\frac{3/2}{4/3}-1)}$

$\Rightarrow fl=fa\times 0.5\times 8=20\times 4=80cm$

charge $=fl-fa=80-20=60cm$