58 through 67 61 59 Lenses with given radii. Objectstands in front of a thin lens, on the central axis. For this situation, each problem in Table 34-7 gives object distance, index of refraction n of the lens, radiusof the nearer lens surface, and radius of the farther lens surface. (All distances are in centimetres.) Find (a) the image distanceand (b) the lateral magnificationof the object, including signs. Also, determine whether the image is (c) realor virtual, (d) invertedfrom object or non-inverted, and (e) on the same side of the lens as objector on the opposite side.

1. The object distance,$P=+60cm$
2. The index of refraction of the lens,$n=1.50$
3. The radius of the nearer lens surface,$r_1=+35cm$
4. The radius of the farther lens surface,$r_2=-35cm$

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 the 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 the same side as the object or on the opposite side.

(a)

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

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

$$\begin{gathered} =\frac{\left(+35cm\right)\left(-35cm\right)}{\left(1.50-1\right)\left(-35cm\right)-\left(+35cm\right)} \\ =+35cm \end{gathered}$$

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

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:

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

$$\begin{gathered} =\frac{\left(+60cm\right)\left(+35cm\right)}{\left(+60cm\right)-\left(+35cm\right)} \\ =+84cm \end{gathered}$$

Hence, the value of the image distance is+84cm.

(b)

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:

$$\begin{gathered} m=\frac{\left(+84cm\right)}{+60cm} \\ =-1.4 \end{gathered}$$

Hence, the value of the lateral magnification is$-1.4$

(c)

The value of image distance is positive.

Hence, the image is real$\left(R\right)$.

(d)

The value of lateral magnification is negative.

Hence, the image is not inverted$\left(NI\right)$.

(e)

The value of image distance is positive and the image is real.

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