Question

An object is placed $10cm$ away from a glass piece $(\mu =1.5)$ of length $20cm$ bound by spherical surfaces of radii of curvature $10cm$. Find the position of final image formed after refractions.

Answer 1

It is given that for first surface:

$n_1=1$

$n_2=1.5$

$v_1=?$

$u_1=-10cm$

$R=+10cm$

So,

$\frac{n_2}{v_1}-\frac{n_1}{u_1}=\frac{n_2-n_1}{R}..........\left(1\right)$

$\frac{1.5}{v_1}-\frac{1}{(-10)}=\frac{1.5-1}{(+10)}$

$v_1=-30cm$

Image is formed at $v_1=-30cm$ away from A.

For second surface:

$n_1=1.5$

$n_2=1$

$u_2=-50cm$

$R=-10cm$

$v_2=?$

$u_2=-10cm$

Using equation (1)

$\frac{1}{v_2}-\frac{1.5}{(-50)}=\frac{1-1.5}{(-10)}$

$v_2=+50cm$

Image is formed at $v_2=+50cm$ away from B.