32 through 38 37, 38 33, 35 Spherical refracting surfaces. An object $\mathbf{\text{O}}$standson the central axis of a spherical refracting surface. For this situation, each problem in Table 34-5 refers to the index of refraction${\mathbf{\text{n}}}_{\mathbf{1}}$where the objectis located, (a) the index of refraction ${\mathbf{n}}_{\mathbf{2}}$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 $\left(R\right)$or virtual $\left(V\right)$and (f) on the same side of the surface asthe object Oor on the opposite side.

• The index of refraction where the object is located is $n_1=1.5$.
• The object distance, $p=+100$
• The image distance,$i=+600$
• The radius of curvature, $r=-30$.

The index of refraction of object, the object distance, the radius of curvature, and the image distance are given in the problem. Using this data and equation 34-8, find the index of refraction on the other side of the refracting surface and check whether the image is real or virtual and find the position of the image.

Formulae are as follows:

$\frac{n_1}{p}+\frac{n_2}{i}=\frac{n_2-n_1}{r}$, where, m is the magnification, p is the pole, and f is the focal length.

From equation 34-8,

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

Rearranging the terms,

$\left(\frac{n_1}{p}+\frac{n_1}{r}\right)=\left(\frac{1}{r}-\frac{1}{i}\right)n_2$

$n_2=\frac{\left(\frac{n}{p}+\frac{n}{r}\right)}{\left(\frac{1}{r}-\frac{1}{i}\right)}$

Substituting the given values,

$n_2=\frac{(\frac{1.5}{100}}{+}\frac{1.5}{\left(\frac{1}{30}\right)}-0-\frac{1}{4600}$

$n_2=1$

Therefore, the index of refraction $n_2$on the other side of the refracting surface is 1.

The object distance is given in the problem, $p=+100cm$

Therefore, the object distance p is +100 cm.

The radius of curvature is given in the problem, r=-30 cm

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

The image distance is given in the problem, i=+600 cm

Therefore, the image distance i is +600 cm.

Since $i>0$, therefore the image is real and inverted.

Therefore, the image is real and inverted.

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 real, therefore theimage is on theoppositeside as that of the object.

Therefore, the image is on the opposite side 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.