91Ó°ÊÓ

$$\begin{gathered} \mathrm{f}=20\mathrm{cm} \\ =0.2\mathrm{m} \end{gathered}$$

• Object distance,$\mathrm{p}=1.5×{10}^{11}\mathrm{m}$
• Radius of the curvature of the sun,${\mathrm{r}}_{\mathrm{s}}=6.96×{10}^8\mathrm{m}$

We know the relation between the magnitude of magnification, image distance and object distance. We can write the relation for the magnitude of magnification, image diameter and object diameter. Using these relations, we can find the diameter of the image.

The expression for the magnification is given by,

$$\begin{gathered} \mathrm{m}=-\frac{\mathrm{i}}{\mathrm{p}} \\ \mathrm{m}=\frac{{\mathrm{d}}_{\mathrm{i}}}{{\mathrm{d}}_{\mathrm{s}}} \end{gathered}$$

Herei is the image distance, p is the object distance, m is the magnification.

From the expression of the magnification,

$\mathrm{m}=-\frac{\mathrm{i}}{\mathrm{p}}$

Considering the magnitude of magnification,

$\left|\mathrm{m}\right|=\frac{\mathrm{i}}{\mathrm{p}}\left(1\right)$

The magnitude of magnification in terms of the diameter can be written as

$\left|\mathrm{m}\right|=\frac{{\mathrm{d}}_{\mathrm{i}}}{{\mathrm{d}}_{\mathrm{s}}}\left(2\right)$

From equations 1 and 2,

$\frac{{\mathrm{d}}_{\mathrm{i}}}{{\mathrm{d}}_{\mathrm{s}}}=\frac{\mathrm{i}}{\mathrm{p}}$

Now, since the rays are coming from infinity, they are parallel to the principal axis.

Therefore, we can say that after the refraction they will meet at the focus of the lens.

So,

i = f

Therefore,

$$\begin{gathered} \frac{{\mathrm{d}}_{\mathrm{i}}}{{\mathrm{d}}_{\mathrm{s}}}=\frac{\mathrm{f}}{\mathrm{p}} \\ {\mathrm{d}}_{\mathrm{i}}={\mathrm{d}}_{\mathrm{s}}×\frac{\mathrm{f}}{\mathrm{p}} \end{gathered}$$

From Appendix C, we have,

$$\begin{gathered} \mathrm{p}=1.5×{10}^{11}\mathrm{m} \\ {\mathrm{r}}_{\mathrm{s}}=6.96×{10}^8\mathrm{m} \\ {\mathrm{d}}_{\mathrm{i}}=2×6.96×{10}^8×\left(\frac{0.2}{1.5×{10}^{11}}\right) \\ =1.86×{10}^{-3}\mathrm{m} \end{gathered}$$

Therefore the diameter of the image is $1.86×{10}^{-3}\mathrm{m}$.