Question

In Fig. 34-38, a beam of parallel light rays from a laser is incident on a solid transparent sphere of an index of refraction n. (a) If a point image is produced at the back of the sphere, what is the index of refraction of the sphere? (b) What index of refraction, if any, will produce a point image at the center of the sphere?

Short answer

1. If a point image is produced at the back of the sphere, the refractive index of the sphere is 2.00.
2. It is not possible to produce a point image at the center of the sphere.

Step-by-step solution

Step 1: Given data

• The object distance is$\mathrm{p}=\infty$
• Refractive index;${\mathrm{n}}_1={\mathrm{n}}_{\mathrm{air}}=1$

Determining the concept

Using the relation between the index of refraction of object and image, image distance, object distance, and the radius of curvature, given by equation 34-8, find the required answers.

Formulae are as follows:

$\frac{{\mathrm{n}}_1}{\mathrm{p}}+\frac{{\mathrm{n}}_2}{\mathrm{i}}=\frac{{\mathrm{n}}_2-{\mathrm{n}}_1}{\mathrm{r}}$

Here, p is the pole, i is the image distance.

(a) Determining the refractive index of the sphere, if a point image is produced at the back of the sphere.

As the image is produced at the back of the sphere, so,

i = 2r

$\frac{{\mathrm{n}}_1}{\mathrm{p}}+\frac{{\mathrm{n}}_2}{\mathrm{i}}=\frac{{\mathrm{n}}_2-{\mathrm{n}}_1}{\mathrm{r}}$

Substituting the given values,

$\frac{1}{\infty }+\frac{{\mathrm{n}}_2}{2\mathrm{r}}=\frac{{\mathrm{n}}_2-1}{\mathrm{r}}$

$$\begin{gathered} \frac{{\mathrm{n}}_2}{2\mathrm{r}}=\frac{{\mathrm{n}}_2-1}{\mathrm{r}} \\ \frac{{\mathrm{n}}_2}{2}={\mathrm{n}}_2-1 \\ {\mathrm{n}}_2\left(1-\frac{1}{2}\right)=1 \\ \frac{{\mathrm{n}}_2}{2}=1 \\ {\mathrm{n}}_2=2.00 \end{gathered}$$

Hence, if a point image is produced at the back of the sphere, the refractive index of the sphere is 2.00.

(b) Determine the refractive index of the sphere that will produce a point image at the center of the sphere.

For an image to be produced at the center of the sphere,

$$\begin{gathered} \mathrm{i}=\mathrm{r} \\ \frac{{\mathrm{n}}_1}{\mathrm{p}}+\frac{{\mathrm{n}}_2}{\mathrm{i}}=\frac{{\mathrm{n}}_2-{\mathrm{n}}_1}{\mathrm{r}} \end{gathered}$$

Substituting the given values,

role="math" localid="1662977840109" $$\begin{gathered} \frac{{\mathrm{n}}_1}{\infty }+\frac{{\mathrm{n}}_2}{\mathrm{r}}=\frac{{\mathrm{n}}_2-1}{\mathrm{r}} \\ \frac{{\mathrm{n}}_2}{\mathrm{r}}=\frac{{\mathrm{n}}_2-1}{\mathrm{r}} \end{gathered}$$

This is not valid unless ${\mathrm{n}}_2\to \infty$or $\mathrm{r}\to \infty$

Hence, it is impossible to produce a point image at the center of the sphere.

The required quantities can be found by using the relation between the index of refraction of object and image, the image distance, the object distance, and the radius of curvature.