91影视

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

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.

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,

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

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

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

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" $$\begin{gathered} \frac{{\mathrm{n}}_1}{鈭�}+\frac{{\mathrm{n}}_2}{\mathrm{r}}=\frac{{\mathrm{n}}_2-1}{\mathrm{r}} \\ \frac{{\mathrm{n}}_2}{\mathrm{r}}=\frac{{\mathrm{n}}_2-1}{\mathrm{r}} \end{gathered}$$

This is not valid unless ${\mathrm{n}}_2鈫�鈭�$or $\mathrm{r}鈫�鈭�$

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.