91Ó°ÊÓ

Object distance, p = 70 cm

Horizontal scale, p= -40 cm

By using the thin lens equation, given by equation 34-9, first you can find the focal length and then the image distance.

Formula:

The lens equation is as follows.

$\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{i}}=\frac{1}{\mathrm{f}}$ ….. (1)

Where, f is the focal length, i is the image distance, and is the object distance.

Rewrite the thin lens equation as below.

$\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{i}}=\frac{1}{\mathrm{f}}$

From the given graph, At the object distance,

role="math" localid="1663223950187" $$\begin{gathered} \mathrm{p}=\frac{{\mathrm{p}}_{\mathrm{s}}}{2} \\ =20\mathrm{cm} \end{gathered}$$

The image distance, i = -10 cm

Therefore, the focal length of the thin lens is,

$$\begin{gathered} \frac{1}{\mathrm{f}}=\frac{1}{20\text{cm}}+\frac{1}{-10\text{cm}} \\ =\frac{\left(-10\text{cm}\right)+20\text{cm}}{\left(-\text{10 cm}\right)\left(\text{20 cm}\right)} \\ \mathrm{f}=\frac{-{\text{200 cm}}^2}{\text{10 cm}} \\ =-20\text{cm} \end{gathered}$$

Hence, the focal length of the thin lens is - 20 cm.

Since the focal length is constant for the given graph. Therefore, the distance of the object is,

p = 70 cm

Write the lens formula again to define the image distance.

$$\begin{gathered} \frac{1}{\mathrm{p}}+\frac{1}{\mathrm{i}}=\frac{1}{\mathrm{f}} \\ \frac{1}{70\text{cm}}+\frac{1}{\mathrm{i}}=\frac{1}{-20\text{cm}} \\ \frac{1}{\mathrm{i}}=\frac{1}{-20\text{cm}}-\frac{1}{70\text{cm}} \\ \frac{1}{\mathrm{i}}=\frac{70\text{cm}-\left(-20\text{cm}\right)}{\left(70\text{cm}\right)\left(-20\text{cm}\right)} \\ \mathrm{i}=\frac{-1400\text{cm}}{90\text{cm}} \\ =-15.56\text{cm} \\ ≈-16cm \end{gathered}$$

Hence, the image distance when p = 70 cm is -16 cm.