It has been recently suggested that the 17th century Dutch artist Rembrandt used a concave mirror to help him etch self-portraits by projecting an inverted image of himself onto a copper sheet - Leaving Cert Physics - Question c - 2017

To find the distance from the sheet to the mirror, we can use the mirror formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

where:

• $f$ is the focal length of the mirror (60 cm).
• $u$ is the distance from the object to the mirror.
• $v$ is the distance from the mirror to the image.

Given that the image is half the size of the object, we have: $m = \frac{h_{image}}{h_{object}} = \frac{1}{2}$

This means: $v = 2u$

Substituting $v$ into the mirror formula gives: $\frac{1}{60} = \frac{1}{u} + \frac{1}{2u}$

Combining the terms: $\frac{1}{60} = \frac{3}{2u}$

Solving for $u$: $u = 90 \text{ cm}$

Thus, $v$ will be: $v = 2u = 180 \text{ cm}$

The distance from the sheet to the mirror is therefore 90 cm.

As calculated above in part (i), the distance from the object to the mirror is:

$u = 90 \text{ cm}$

A concave mirror can produce an upright image; however, this is typically only valid when the object is located within the focal length. Since Rembrandt was using the mirror to project an inverted image onto the copper sheet to etch self-portraits, a virtual, upright image could not be formed on the sheet. Therefore, this image type would not assist him in creating accurate depictions, as it did not allow for the necessary projection to be captured as intended.