Question

50 through 57 55, 57 53 Thin lenses. Object $\mathbf{O}$stands on the central axis of a thin symmetric lens. For this situation, each problem in Table 34-6 gives object distance p (centimeters), the type of lens (C stands for converging and D for diverging), and then the distance (centimeters, without proper sign) between a focal point and the lens. Find (a) the image distance $\mathbf{i}$ and (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real $\left(R\right)$ or virtual $\left(V\right)$ , (d) inverted $\left(I\right)$ from object $\mathbf{O}$ or non inverted $\left(\mathrm{NI}\right)$, and (e) on the same side of the lens as object $\mathbf{O}$or on the opposite side.

Short answer

1. Image distance$\mathrm{i}=-4.8\mathrm{cm}$
2. Lateral magnification$\mathrm{m}=+0.60$
3. Image is virtual$\left(\mathrm{V}\right)$
4. Image is non-inverted$\left(\mathrm{NI}\right)$
5. Image is on the same side of the object.

Step-by-step solution

Listing the given quantities

The lens is diverging

Focal length,$\mathrm{f}=-12.0\mathrm{cm}$

Object distance, $\mathrm{p}=+8$

Understanding the concepts of lens equation and the formula for magnification

By using the thin lens equation and the formula for magnification, we can find all the required quantities.

Formula:

Thin lens equation,$\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{i}}$

Magnification,$\mathrm{m}=-\frac{\mathrm{i}}{\mathrm{p}}$

(a) Calculations of the image distance

Since the lens is diverging, the focal length value should be negative, i.e.

$\mathrm{f}=-12\mathrm{cm}$

Thin lens equation is

$\begin{array}{c}\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{i}}\\ \frac{1}{(-12)}=\frac{1}{8}+\frac{1}{\mathrm{i}}\\ \frac{1}{\mathrm{i}}=\frac{1}{(-12)}-\frac{1}{8}\\ \frac{1}{\mathrm{i}}=-0.2083\end{array}$

$\mathrm{i}=-4.8\mathrm{cm}$

Image distance $\mathrm{i}=-4.8\mathrm{cm}$

(b) Calculations of the magnification 

Magnification is,

$\begin{array}{c}\mathrm{m}=-\frac{\mathrm{i}}{\mathrm{p}}\\ =-\left(\frac{-4.8}{8}\right)\\ =+0.60\end{array}$

Lateral magnification $\mathrm{m}=+0.60$

(c) Explanation

As the image distance$\mathrm{i}$ is negative, the image is virtual $\left(\mathrm{V}\right)$.

(d) Explanation

As the magnification is positive, the image is non-inverted $\left(NI\right)$.

(e) Explanation 

For thin lens, the real images form on the opposite side as the object and virtual images form on the same side as the object.

Since the image is non-inverted, it forms on the same side of the object.