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 $\mathit{r}\mathbf{1}$ of the nearer lens surface, and radius $\mathit{r}\mathbf{2}$of the farther lens surface. (All distances are incentimeters.) 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 noninverted (NI), and (e) on the same side of the lens as object $\mathit{O}$or on the opposite side.

The object distance is$P=+29\mathrm{cm}$

The index of refraction of the lens is$n=1.65$

The radius of the nearer lens surface is

$r_1=+35\mathrm{cm}$

The radius of the farther lens surface is $r_2=\infty$

We can use the concept of the Lens 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 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:

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

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

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

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

$$\begin{gathered} \frac{1}{f}=\left(n-1\right)\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \\ =\left(1.65-1\right)\left(\frac{1}{35cm}-\frac{1}{\infty }\right) \\ =1.9\times {10}^{-2}cm \end{gathered}$$

$f=+54\mathrm{cm}$

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 $i$ are related to the lens’s focal length.

$$\begin{gathered} \frac{1}{f}=\frac{1}{P}+\frac{1}{i} \\ \frac{1}{i}=\frac{1}{f}-\frac{1}{P} \end{gathered}$$

$$\begin{gathered} i=\frac{Pf}{P-f} \\ =\frac{\left(+29cm\right)\left(+54cm\right)}{\left(+29cm\right)-\left(+54cm\right)} \\ =-63cm \end{gathered}$$

The image distance $i=-63\mathrm{cm}$

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

$$\begin{gathered} m=-\frac{i}{P} \\ =-\frac{\left(-63cm\right)}{+29cm} \\ =+2.2 \end{gathered}$$

The lateral magnification of the object is $m=+2.2$

Whetherthe image is real $\left(R\right)$or virtual$\left(V\right)$:

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$\left(V\right)$ .

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