In a concave mirror, an object is placed at a distance x from the focus and the image is formed at a distance y from the focus. What is the focal length of the mirror?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( (xy)^{\frac{1}{2}} \)
When the distances are measured from the focus instead of the pole, the mirror formula takes the form \( f^{2} = xy \) (Newton's formula). It follows from the mirror formula with the Cartesian sign convention (distances measured against the incident light are negative).
For a real object and a real image in front of the mirror:
Let \( f \) be the magnitude of the focal length, so the focal length is \( -f \), and
\( u = -(f + x) \)
\( v = -(f + y) \)
\( \displaystyle \frac{1}{v} + \frac{1}{u} = \frac{1}{-f} \;\Rightarrow\; -\frac{1}{f + y} - \frac{1}{f + x} = -\frac{1}{f} \)
\( \displaystyle f\,(2f + x + y) = (f + x)(f + y) \)
\( \displaystyle 2f^{2} + fx + fy = f^{2} + fx + fy + xy \)
\( \displaystyle f^{2} = xy \;\Rightarrow\; f = (xy)^{\frac{1}{2}} \)
Shortcut: Dimensions rule out two options at once: \( xy \) is a length squared and \( xy^{-1} \) has no unit, so neither can be a focal length. Putting the object at the centre of curvature gives \( x = y = f \), and \( (x + y)/2 = f \) as well, but at any other position, for example \( x = f/2 \) and \( y = 2f \), only \( (xy)^{1/2} \) still gives \( f \).