32 through 38 37, 38 33, 35 Spherical refracting surfaces. An object $\mathit{O}$stands on the central axis of a spherical refracting surface. For this situation, each problem in Table 34-5 refers to the index of refraction ${\mathit{n}}_{\mathbf{1}}$where the object is located, (a) the index of refraction ${\mathit{n}}_{\mathbf{2}}$on the other side of the refracting surface, (b) the object distance$\mathit{p}$, (c) the radius of curvature $\mathit{r}$of the surface, and (d) the image distance $\mathit{i}$. (All distances are in centimeters.) Fill in the missing information, including whether the image is (e) real $\mathbf{\left(}\mathbf{R}\mathbf{\right)}$or virtual $\mathbf{\left(}\mathbf{V}\mathbf{\right)}$and (f) on the same side of the surface as the object $\mathit{O}$or on the opposite side.

• The index of refraction where the object is located is $n_1=1.0$.
• The index of refractionon the other side of the refracting surface,$n_2=1.5$.
• Object distance,$p=+10$.
• The radius of curvature, $r=+30$.

The index of refraction of object and image, object distance, and radius of curvature are given in the problem. Using this data and equation, $\mathbf{34}\mathbf{-}\mathbf{8}$,find theimage distance and check whetherthe image is real or virtual, and find the position of the image.

Formulae are as follows:

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

Here, $\mathit{p}$is the pole, $\mathit{i}$is the image distance.

The index of refractionon the other side of the refracting surfaceis given in the table$34-5$. So,$n_2=1.5$

Therefore, the index of refraction $n_2$on the other side of the refracting surface is $1.5$.

Theobject distance is given in the problem,$p=+10\mathrm{cm}$.

Therefore, the object distance $p$is $+10\mathrm{cm}$.

The radius of curvature is given in the problem, $r=+30cm$.

Therefore, the radius of curvature $r$of the surface is $+30\mathrm{cm}$.

Fromequation,$34-8$,
$\frac{n_1}{p}+\frac{n_2}{i}=\frac{n_2-n_1}{r}$

Rearranging the terms,

$i=\frac{n_2}{\left(\frac{n_2-n_1}{r}-\frac{n_1}{p}\right)}$

Substituting the given values,

$i=\frac{\text{1.5}}{\left(\frac{\text{1.5 - 1.0}}{\text{30cm}}\text{-}\frac{\text{1.0}}{\text{10cm}}\right)}$

$i=-18cm$

Therefore, the image distance $i$is $-18\mathrm{cm}$.

Since $i<0$, therefore the image is virtual and upright.

Therefore, the image is virtual and upright.

For spherical refracting surfaces, real images form on the opposite sides of the object and virtual images form on the same side as the object.

Since the image is virtual, therefore theimage is on the same side as that of object.

Therefore, the image is on the same side as that of the object.

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