Question

A converging lens with a focal length of 12.0 cmforms a virtual image 8.00 mmtall, 17.0 cmto the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.

Short answer

The position and size of the object is 7 cm and 3.31 mm .

The image formed is erect and virtual, thus, the object and the image are in the same side of the lens.

Step-by-step solution

Determine the position and size of the object

Now, here as the image is virtual, thus s' is negative

Use the object- image relationship

$\frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{f}$

Substitute the given values to find the required answer

$$\begin{gathered} s=\frac{fs\text{'}}{s\text{'}-f} \\ s=\frac{\left(12cm\right)\left(-17cm\right)}{-17cm-12cm} \\ \Rightarrow s=7cm \end{gathered}$$

Now, use the formula of lateral magnification for a thin less

$$\begin{gathered} m=\frac{-s\text{'}}{s}=\frac{y\text{'}}{y} \\ m=\frac{-s\text{'}}{s}=\frac{-\left(-17cm\right)}{7cm}=+2.41667 \\ y=\frac{y\text{'}}{m}=\frac{8mm}{2.41667}=3.31mm \end{gathered}$$

Thus, the image is virtual, so the object and the image are in the same side of the lens.

Therefore, the required diagram is