91影视

• Radius of curvature, r = -30 cm
• The image distance, i = -7.5 cm

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

Formulae are as follows:

$\frac{{\mathbf{n}}_{\mathbf{1}}}{\mathbf{p}}\mathbf{+}\frac{{\mathbf{n}}_{\mathbf{2}}}{\mathbf{i}}\mathbf{=}\frac{{\mathbf{n}}_{\mathbf{2}}\mathbf{-}{\mathbf{n}}_{\mathbf{1}}}{\mathbf{r}}$, where p is the pole, and i is the image distance.

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

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

Rearranging the terms,

$$\begin{gathered} \frac{n_1}{p}=\frac{n_2-n_1}{r}-\frac{n_2}{i} \\ p=\frac{n_1}{\frac{n_2-n_1}{r}-\frac{n_2}{i}} \end{gathered}$$

Substituting the given values,

$$\begin{gathered} p=\frac{1.5}{\frac{1.0-1.5}{-30}-\frac{1.0}{-7.5}} \\ p=+10\text{cm} \end{gathered}$$

Therefore, the object distance p is +10 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 = -7.5 cm.

Therefore, the image distance i is -7.5 cm.

Since, i < 0, therefore the image 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 the object.

Therefore, the image and object are on the same side.

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