Question

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

Short answer

a) The index of refraction $n_2$on the other side of the refracting surface is 1.0.

b) The object distance p is +10 cm.

c) The radius of curvature r of the surface is 30 cm.

d) The image distance i is -6 cm.

e) The image is virtual and upright.

f) The image is on same side as that of the object.

Step-by-step solution

Step 1: Given

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

Table 34-5

Determining the concept

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

Determining the index of refraction  on the other side of the refracting surface

(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.

Determining the object distance 

(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.

Determining the radius of curvature r of the surface

(c)

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

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

Determining the image distance i

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.

Determining whether the image is real or virtual

(e)

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

Therefore, the image is virtual and upright.

Determining the position of the image

(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.