32 through 38 37, 38 33, 35 Spherical refracting surfaces. An object Ostands on the central axis of a sphericalrefractingsurface. 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 p, (c) the radius of curvature rof the surface, and (d) the image distance i. (All distances are in centimeters.) Fill in the missing information, including whether the image is (e) real (R)or virtual (V)and (f) on the same side of the surface as object Oor on the opposite side

$$\begin{gathered} n_1=1.5 \\ {n}_2=1.0 \\ p=+10.0cm \\ i=-6.0 \end{gathered}$$

Table 34-5

The index of refraction of object and image, the object distance and the image distance are given i the problem. Using this data and equation, find the radius of curvature and check whether the image is real or virtual and find the position of the image.

Formulae are as follows:

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

where p is the pole and i is the image distance

(a)

Index of refraction on the other side of the refracting surface is given in the table 34-5. So $n_2=1.0$.

Therefore, the index of refraction $n_2$on the other side of the refracting surface is 1.0.

(b)

The index of refraction on the other side of the refracting surface is given in the table 34-5. So, $n_2=1.0$.

Therefore, the object distance p is +10 cm.

(c)

Theobject distance is given in the problem, p = +10 cm

Therefore, the radius of curvature r of the surface is 30 cm.

d)

From equation 34-8

$\frac{n_1}{p}+\frac{n_2}{i}=\frac{n_2-n_1}{r}$

Rearranging the terms,

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

Substituting the given values

role="math" localid="1663044974048" $$\begin{gathered} r=\frac{\left(1.0-1.5\right)}{\left(\frac{1.5}{10}+\frac{1.0}{6}\right)} \\ r=30\text{cm} \end{gathered}$$

Therefore, the image distance i is -6 cm.

(e)

The image distance is given in the problem i = -6 cm

Therefore, the image is virtual and upright.

(f)

For spherical refracting surfaces, real images form on the opposite side 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 the object.

Therefore, the image is on 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.