Question

A hemispherical portion of the surface of a solid glass sphere (μ = 1.5) of radius r is silvered to make the inner side reflecting. An object is placed on the axis of the hemisphere at a distance 3r from the centre of the sphere. The light from the object is refracted at the unsilvered part, then reflected from the silvered part and again refracted at the unsilvered part. Locate the final image formed.

Answer 1

The final image is formed on the reflecting surface of the sphere.

Explanation:

Given data in the question  :

A hemispheric portion of a solid glass sphere's surface is $\mu_{1}, \mu_{2}$.

$\mu_{2}=1.5$

$\mu_{1}=1$

u = 2r      

r  is radius  

We know that ,

                    $\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}$

                 $\frac{1.5}{v}-\frac{1}{-2 r}=\frac{1.5-1}{r}$

                  $\frac{1.5}{v}-\frac{1}{-2 r}=\frac{0.5}{r}$

                  $\frac{1.5}{v}-\frac{1}{-2 r}=\frac{5}{10 r}$

                  $\frac{1.5}{v}-\frac{1}{-2 r}=\frac{1}{2 r}$

                     $\frac{1.5}{v}+\frac{1}{2 r}=\frac{1}{2 r}$

V = ∞

A concave mirror reflection

u = ∞

So that, v = focal length of mirror = $\frac{r}{2}$

Here,

                    u = $\frac{-3 r}{2}$

                    R = -r

         $\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}$

         $\frac{1}{v}-\frac{1.5}{\frac{-3 r}{2}}=\frac{-0.5}{-r}$

     $\frac{1}{v}-\frac{1.5 \times 2}{-3 r}=\frac{-0.5}{-r}$

         $\frac{1}{v}-\frac{3}{-3 r}=\frac{-0.5}{-r}$

           $\frac{1}{v}-\frac{1}{-r}=\frac{-0.5}{-r}$

              $\frac{1}{v}+\frac{1}{r}=\frac{0.5}{r}$

              $\frac{1}{v}+\frac{1}{r}=\frac{5}{10 r}$

              $\frac{1}{v}+\frac{1}{r}=\frac{1}{2 r}$

                     $\frac{1}{v}=\frac{1}{2 r}-\frac{1}{r}$

                     $\frac{1}{v}=\frac{1-2}{2 r}$

                     $\frac{1}{v}=\frac{-1}{2 r}$

                     -v = 2r

               v = -2r

Here, negative sign represents that image is formed inside APB, but the image formation must be at C.

Thus, the final image is formed.