91Ó°ÊÓ

$n_1=1$for air and $n_2=2$

Here, we need to use the equation of spherical refracting surface. As the rays are parallel to the surface, we can assume the object's distance at infinity. Hence, by solving that equation, we can get the image distance. This image will act as the object for the second refracting surface. So, using the equation for the second refracting surface, we can get an expression for the final image distance.

Formula:

For spherical refracting surface

$\frac{{\mathbf{n}}_{\mathbf{1}}}{\mathbf{p}}\mathbf{+}\frac{{\mathbf{n}}_{\mathbf{2}}}{\mathbf{i}}\mathbf{=}\frac{\mathbf{(}{\mathbf{n}}_{\mathbf{2}}\mathbf{-}{\mathbf{n}}_{\mathbf{1}}\mathbf{)}}{\mathbf{r}}$ . . . (34-8)

(a)

For spherical refracting surface,

$\frac{n_1}{p}+\frac{n_2}{i}=\frac{(n_2-n_1)}{r}$

Let the object be placed at infinity as the rays are arriving parallel, then we set $p→∞$in the above equation, we get

$\begin{array}{l}\frac{n_2}{i}=\frac{(n_2-n_1)}{r}\\ i=\frac{n_{2}r}{n_2-n_1}\end{array}$

Now set $n_2=n$and $n_1=1$we get

$i=\frac{n·r}{n-1}$

This image will serve as the virtual object for the second imaging event so the distance of the virtual object is

$\begin{array}{c}p\text{'}=2r-i\\ =2r-\frac{n·r}{n-1}\\ =\frac{r\left(n-2\right)}{n-1}\end{array}$

Let $p\text{'}$be the new distance for the object and $i\text{'}$be the new distance for the image and since the right part of the sphere is concave $r<0$using the formula for the spherical refracting mirror, we get

For the following calculation, we will set

$n_1=n$; $n_2=1andr<0$

$\begin{array}{c}\frac{n}{p\text{'}}+\frac{1}{i\text{'}}=\frac{\left(-1+n\right)}{r}\\ =\frac{n-1}{r}\end{array}$

Substituting $p\text{'}$in the above equation, we get

$\begin{array}{c}\frac{1}{i\text{'}}=\frac{n-1}{r}-\frac{n}{p\text{'}}\\ =-\frac{2\left(n-1\right)}{r\left(n-2\right)}\\ =\frac{2\left(n-1\right)}{r\left(2-n\right)}\end{array}$

The final image position is

$i\text{'}=\frac{r\left(2-n\right)}{2\left(n-1\right)}$

(b)

$1<n<2$and$i\text{'}>0$

The image will be at the right of the right side of the sphere