Discussion Forum

The diameter pf a plano-convex lens is 6 cm and thickness at the centre is 3mm. If the speed of light in material of lens is $2\times 10^{8}$  m/s, the  focal length of lens is 

Answer:

Explanation:

By pythagoras theorm

$R^{2}=(3)^{2}+(R-0.3)^{2}$

   = R=15 cm

  Refractive index of material of lens

               $\mu=\frac{c}{v}$

Here , c= speed of light in vacum    =3 x10⁸  m/s

 v= speed of light in material of lens  = 2 x 10⁸ m/s

  =   $\frac{3\times 10^{8}}{2 \times 10^{8}}=\frac{3}{2}$

 From lens marker's formula

       $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

 Here, R₁=R and R₂= $\infty$ (For plane surface)

$\frac{1}{f}=(\frac{3}{2}-1)\left(\frac{1}{15}\right)$

  = f= 30 cm