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 $O$, index of refraction n of the lens, radius of the nearer lens surface, and radius of the farther lens surface. (All distances are in centimeters.) Find (a) the image distance and (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real or virtual , (d) inverted from the object $O$or non-inverted , and (e) on the same side of the lens as object or on the opposite side.

• The object distance,$P=+10\text{cm}$
• The index of refraction of the lens,$n=1.50$
• The radius of the nearer lens surface,$r_1=+30\text{cm}$
• The radius of the farther lens surface,$r_2=-30\text{cm}$

We can use the concept of the lens formula and the 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 image as well as a real one. 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.

Formulae:

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, $\frac{1}{f}=\frac{1}{P}+\frac{1}{i}$(ii)

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

Here $f$ is the focal length, $i$is the image distance, $p$ is the object distance, $m$ is the magnification.

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

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

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

The given lens is a converging lens because its focal length is positive.

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

$$\begin{gathered} \frac{1}{i}=\frac{1}{f}-\frac{1}{P} \\ i=\frac{Pf}{P-f} \\ =\frac{\left(+10\text{cm}\right)\left(+30\text{cm}\right)}{\left(+10\text{cm}\right)-\left(+30\text{cm}\right)} \\ =-\text{15cm} \end{gathered}$$

Hence, the value of the image distance is$-\text{15cm}$ .

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

$$\begin{gathered} m=-\frac{\left(-15\text{cm}\right)}{+10\text{cm}} \\ =+1.5 \end{gathered}$$

Hence, the value of the lateral magnification is $+1.5$.

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.