Question

A narrow beam of parallel light rays is incident on a glass sphere from the left, directed toward the center of the sphere. (The sphere is a lens but certainly not a thin lens.) Approximate the angle of incidence of the rays as $\mathbf{0}\mathbf{°}$, and assume that the index of refraction of the glass is $\mathit{n}\mathbf{}\mathbf{<}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}$(a) In terms of n and the sphere radius $\mathit{r}$, what is the distance between the image produced by the sphere and the right side of the sphere? (b) Is the image to the left or right of that side? (Hint: Apply Eq. 34-8 to locate the image that is produced by refraction at the left side of the sphere; then use that image as the object for refraction at the right side of the sphere to locate the final image. In the second refraction, is the object distance positive or negative?)

Short answer

Distance between the right side of the sphere and the image of the object$i\text{'}=\frac{r\left(2-n\right)}{2\left(n-1\right)}$

The location of the image is at the right of the right side of the sphere.

Step-by-step solution

Listing the given quantities

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

Understanding the equations of surface

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)

calculations of the distance between the right side of the sphere and the image of the object

(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 $p\to \infty$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

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

location of the image

(b)

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

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