Question

An object of height \(h\) is placed at a distance \(d_{\mathrm{a}}\) on the left side of a converging lens of focal length \(f\left(f

Short answer

Answer: The object should be placed at a distance of \(d_o = \frac{2f}{3}\) from the converging lens, and the magnification of the system is \(M = -\frac{9}{2}\).

Step-by-step solution

Write down the lens formula

The lens formula is given by \[\frac{1}{f} = \frac{1}{d_i} + \frac{1}{d_o}\] where \(f\) is the focal length, \(d_i\) is the image distance, and \(d_o\) is the object distance.

Plug in the given values and solve for \(d_o\)

We are given that the image distance is \(3f\), so we can substitute that value into the lens formula: \[\frac{1}{f} = \frac{1}{3f} + \frac{1}{d_o}\] To solve for \(d_o\), we first calculate the difference of the fractions on the right side: \[\frac{1}{f} = \frac{2}{3f}\] Next, we cross-multiply the equation: \[2f = 3d_o\] Finally, we solve for \(d_o\): \[d_o = \frac{2f}{3}\]

Calculate the magnification

The magnification formula for lenses is given by \[M = \frac{h_i}{h_o} = -\frac{d_i}{d_o}\] We will now find the values for \(M\) using the given information and the value we've found for \(d_o\). We are given that the image is formed on the right side of the lens, which means that \(d_i\) is positive, so the magnification will be negative because the image is inverted. We know \(d_i = 3f\) and \(d_o = \frac{2f}{3}\). Now, we substitute these values in the magnification formula: \[M = -\frac{3f}{\frac{2f}{3}}\] \[M = -\frac{9}{2}\] So, the distance of the object required to form an image at \(3f\) on the right side of the lens is \(d_o = \frac{2f}{3}\) and the magnification of the system is \(M = -\frac{9}{2}\).