Question

A glass sphere has radius r=-50 cmand index of refraction ${\mathbf{n}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}$paperweight is constructed by slicing through the sphere along a plane that is 2.0 cmfrom the center of the sphere, leaving height p = h = 3.0 cm. The paperweight is placed on a table and viewed from directly above by an observer who is distance d=8.0 cmfrom the tabletop (Fig. 34-39). When viewed through the paperweight, how far away does the tabletop appear to be to the observer?

Short answer

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

Step-by-step solution

Step 1: Given

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

Determining the concept

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.

Determining how far away the tabletop appears to the observer 

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" $=7.42\text{cm}\approx 7.4\text{cm}$

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.