58 through 67 61 59 Lenses with given radii. Object $\mathit{O}$stands in front of a thin lens, on the central axis. For this situation, each problem in Table 34-7 gives object distance $\mathit{p}$ , index of refraction n of the lens, radius localid="1663142386474" ${\mathit{r}}_{\mathbf{1}}$of the nearer lens surface, and radius localid="1663142397129" ${\mathit{r}}_{\mathbf{2}}$of the farther lens surface. (All distances are in centimeters.) Find (a) the image distance $\mathit{i}$and (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real (R) or virtual $\mathbf{\left(}\mathit{V}\mathbf{\right)}$, (d) inverted $\mathbf{\left(}\mathit{I}\mathbf{\right)}$ from object $\mathit{O}$ or non inverted (NI), and (e) on the same side of the lens as object $\mathit{O}$or on the opposite side

The object distance is$P=+6.0\mathrm{cm}$

The index of refraction of the lens is$n=1.70$

The radius of the nearer lens surface is$r_1=+10\mathrm{cm}$

The radius of the farther lens surface is $f_2=-12\mathrm{cm}$

We can use thelens formula and lens marker’s equation. The focal length of the lens is positive for a converging lens and negative for a diverging lens. The converging lens can form a virtual as well as a real image. If the object is outside the focal point, then it is a real image, and if the object is inside the focal point, then it is a virtual image.

Formula:

$\frac{1}{f}=\left(n-1\right)\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$

$\frac{1}{f}=\frac{1}{P}+\frac{1}{i}$

$m=-\frac{i}{P}$

In the given problem, $r_1$is positive and$r_2$is negative; hence the given lens is of double-convex type.

According to the lens marker’s equation, the expression of the focal length of the lens in air is

$$\begin{gathered} \frac{1}{f}=\left(n-1\right)\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \\ \frac{1}{f}=\left(n-1\right)\left(\frac{r_2-r_1}{r_1{r}_2}\right) \end{gathered}$$

$$\begin{gathered} f=\frac{r_1{r}_2}{\left(n-1\right)\left(r_2-r_1\right)} \\ =\frac{\left(+10cm\right)\left(-12cm\right)}{\left(1.70-1\right)\left(-12cm\right)-\left(+10cm\right)} \\ =+7.79cm \end{gathered}$$

The given lens is a converging lens because the focal length ispositive.

For an object in front of the lens, the object distance $P$and image distance are related to the lens focal length.

$\frac{1}{f}=\frac{1}{P}+\frac{1}{i}$

$\frac{1}{i}=\frac{1}{f}-\frac{1}{P}$

$$\begin{gathered} i=\frac{Pf}{P-f} \\ =\frac{\left(+6.0cm\right)\left(+7.79cm\right)}{\left(+6.0cm\right)-\left(+7.79cm\right)} \\ =-26cm \end{gathered}$$

The image distance $i=-26\mathrm{cm}$

The lateral magnification is the ratio of the object distance $P$ to the image distance $i$. It is given by

$$\begin{gathered} m=-\frac{i}{P} \\ =-\frac{\left(-26cm\right)}{+6.0cm} \\ =+4.3 \end{gathered}$$

The lateral magnification of the object is $m=+43$

Whetherthe image is real$R$or virtual$V$:

If the object is inside the focal point, then it is a virtualimage.The image distance is alsonegative; hence the image is virtual$\left(V\right)$.

Whether the image is inverted from object $I$or not inverted $\left(NI\right)$:

The value of magnification is positive; hence the image is not inverted $\left(NI\right)$.

The position of the image:

The image distance is negative; hence the image is on the same side as the object.