Question

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.

Short answer

1. The image distance is $-15\text{cm}$.
2. The lateral magnification of the object is$+1.5$ .
3. The image is virtual $\left(V\right)$.
4. The image is not inverted $\left(NI\right)$from object.
5. The image on the same side of the object.

Step-by-step solution

The data given

• 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,$r_2=-30\text{cm}$

Understanding the concept of properties of the lens

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.

(a) Calculation of the object distance

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$-\text{15cm}$ .

(b) Calculation of the lateral magnification of the object

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 $+1.5$.

(c) Calculation of the behavior of the image

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.

(d) Calculation if the image is inverted or not

The value of magnification is positive.

Hence, the image is not inverted .

(e) Calculation of the position of the image

The image distance is negative.

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