58 through 67 61 59 Lenses with given radii. An object $O$stands in front of a thin lens, on the central axis. For this situation, each problem in Table 34-7 gives object distance $p$, index of refraction n of the lens, radius $r_1$of the nearer lens surface, and radius localid="1663061304344" $r_2$of the farther lens surface. (All distances are in centimeters.) Find (a) the image distance $i$and (b) the lateral magnification $m$of the object, including signs. Also, determine whether the image is (c) real $\left(R\right)$or virtual $\left(V\right)$,(d) inverted $\left(I\right)$from the object $O$or non-inverted $\left(NI\right)$, and (e) on the same side of the lens as object $O$or on the opposite side.

1. The object distance, $P=+10\mathrm{dm}$
2. The index of refraction of the lens,$n=150$
3. The radius of the nearer lens surface, $r_1=-30\mathrm{mm}$
4. The radius of the farther lens surface, localid="1663063178742" $r_2=-60\mathrm{mm}$

Here, we need to use the concept of image formation by the thin lens. We can use the equation 34.9 and 34.10 together to solve for the image distance. The magnification of a lens can be calculated using equation 34.7. By using the values of image distance and magnification, we can determine whether the image is real or virtual, whether it is inverted or non-inverted and whether it is on the same side of the object or the opposite side.

Formula:

The lens formula for refraction, $\frac{1}{f}=\left(n-1\right)\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$ (i)

The lens formula, $\frac{1}{f}=\frac{1}{\overline{P}}+\frac{1}{i}$ (ii)

The magnification formula of the lens, $m=-\frac{i}{P}$ (iii)

Using the given data in equation (i), the expression of the focal length of the lens in air is given as follows:

$$\begin{gathered} f=\frac{r_1{r}_2}{\left(n-1\right)\left(r_2-r_1\right)} \\ =\frac{\left(-30\text{cm}\right)\left(-60\text{cm}\right)}{\left(1.50-1\right)\left(\left(-60\text{cm}\right)-\left(-30\text{cm}\right)\right)} \\ =-120\text{cm} \end{gathered}$$

The given lens is a diverging lens because focal length is negative.

For an object in front of the lens, the object distance and image distance are related to the lens’ focal length. Thus, the image distance can be given using the data in equation (ii) as follows:

$$\begin{gathered} \frac{1}{i}=\frac{1}{f}-\frac{1}{P} \\ i=\frac{Pf}{P-f} \\ =\frac{\left(+10\text{cm}\right)\left(-120\text{cm}\right)}{\left(+10\text{cm}\right)-\left(-120\text{cm}\right)} \\ =-9.2\text{cm} \end{gathered}$$

Hence, the value of the image distance is $-9.2\text{cm}$

The lateral magnification is the ratio of the object distance to the image distance. It is given using the data in equation (iii) as follows:

$$\begin{gathered} m=-\frac{\left(-9.2\text{cm}\right)}{+10\text{cm}} \\ =+0.92 \end{gathered}$$

Hence, the value of the lateral magnification is role="math" localid="1663063492653" $+0.92$

.

If the object is inside the focal point, then it is a virtual image.The image distance is also negative.

Hence, the image is virtual

The value of magnification is positive.

Hence, the image is not inverted.

The image distance is negative.

Hence, the image is on the same side of the object.