91Ó°ÊÓ

$$\begin{gathered} \mathrm{r}=-50\mathrm{cm} \\ {\mathrm{n}}_1=1.6 \\ {\mathrm{n}}_2=1 \\ \mathrm{p}=\mathrm{h}=3.0\text{cm} \\ \mathrm{d}=8.0\text{cm} \end{gathered}$$

Usingthe relationbetweenthe index of the refraction of object and image,the image distance, the object distance, and the radius of curvature, given by equation,find the required answers.

Formula are as follows:

$\frac{{\mathbf{n}}_{\mathbf{1}}}{\mathbf{p}}\mathbf{+}\frac{{\mathbf{n}}_{\mathbf{2}}}{\mathbf{i}}\mathbf{=}\frac{{\mathbf{n}}_{\mathbf{2}}\mathbf{-}{\mathbf{n}}_{\mathbf{1}}}{\mathbf{r}}$

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

Using sign convention, the radius of the sphere is$\mathrm{r}=-5.0\text{cm}$

$$\begin{gathered} \frac{{\mathrm{n}}_1}{\mathrm{p}}+\frac{{\mathrm{n}}_2}{\mathrm{i}}=\frac{{\mathrm{n}}_2-{\mathrm{n}}_1}{\mathrm{r}} \\ \frac{{\mathrm{n}}_2}{\mathrm{i}}=\frac{{\mathrm{n}}_2-{\mathrm{n}}_1}{\mathrm{r}}-\frac{{\mathrm{n}}_1}{\mathrm{p}} \\ \mathrm{i}=\frac{{\mathrm{n}}_2}{\frac{{\mathrm{n}}_2-{\mathrm{n}}_1}{\mathrm{r}}-\frac{{\mathrm{n}}_1}{\mathrm{p}}} \\ \mathrm{i}=\frac{1.0}{\frac{1.0-1.6}{-5.0}-\frac{1.6}{3.0}} \\ \mathrm{i}=-2.42\mathrm{cm} \end{gathered}$$

When viewed through the paperweight, the distance of the tabletop from the observer is$\left(\mathrm{d}-\mathrm{h}\right)+\left|\mathrm{i}\right|$

Distance of tabletop from an observer$=\left(8-3\right)+\left|-2.42\right|=5+2.42$

Distance of tabletop from an observerrole="math" localid="1662979876733" $=7.42\text{cm}≈7.4\text{cm}$

Therefore, the tabletop appears to be at a distance 7.4 cm from the observer.

Using the relation between the index of refraction of object and image, image distance, object distance, and the radius of curvature, the required distance can be found.