91Ó°ÊÓ

1. The object-image separation,$d=40.0cm$
2. The image is inverted and real.
3. The size of image is half of size of object.

Here, we can determine the type of lens from the nature of the image. To find the object distance, we need to use the equation of the magnification given by equations 34.5 and 34.6 and the given value of the object–image separation distance. We can calculate the focal distance using lens equation 34.4.

Since the image formed is real, thus, the lens must be a converging lens.

Hence, it is a type of converging lens.

Using the given data in equation (iii), we can get the magnification value of the object as follows:

$m=\frac{-1}{2}$

Now, the image distance using the above value in equation (ii) can be given as follows:

$$\begin{gathered} \frac{1}{2}=\frac{i}{p} \\ i=\frac{p}{2}..............................\left(a\right) \end{gathered}$$

Now, using the above value and given data that the object-image separation$d=40cm$ , we can get the object distance from the mirror as follows:

$$\begin{gathered} i+p=40.0 \\ \frac{p}{2}+p=40.0 \\ p=\frac{2}{3}×40.0 \\ =26.66667\text{cm} \\ ≈26.7\text{cm} \end{gathered}$$

Hence, the value of the object distance is $26.7cm$.

Substituting the object distance value in equation (a), we can get the image distance as follows:

$$\begin{gathered} i=\frac{26.7\text{cm}}{2} \\ =13.33\text{cm} \end{gathered}$$

Now, using the data in equation (i), we can get the focal length as follows:

$$\begin{gathered} \frac{1}{f}=\frac{1}{13.33}+\frac{1}{26.67} \\ =0.1125 \\ f=8.88889\text{cm} \\ ≈8.89\text{cm} \end{gathered}$$

Hence, the value of focal length is $8.89cm$.