Question

An object of size \(4.0\text{ cm}\) is placed \(25.0\text{ cm}\) in front of a concave mirror of focal length \(15.0\text{ cm}\). At what distance from the mirror should a screen be placed to obtain a sharp image? Determine the nature and size of the image.

Hint:

Always double-check your sign conventions: \(u\) is almost always negative, \(f\) is negative for concave and positive for convex mirrors.

Answer 1

Step 1: Understanding the Concept:
We will use the mirror formula and linear magnification formula applying proper sign conventions.
For a concave mirror, focal length \(f\) is negative. Object distance \(u\) is negative.

Step 2: Key Formula or Approach:
Mirror formula: \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
Magnification formula: \(m = \frac{h'}{h} = -\frac{v}{u}\)

Step 2: Detailed Explanation:
Given:
Object distance, \(u = -25.0\text{ cm}\)
Focal length, \(f = -15.0\text{ cm}\)
Object height, \(h = +4.0\text{ cm}\)
Using the mirror formula to find image distance \(v\):
\[ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \] \[ \frac{1}{v} = \frac{1}{-15.0} - \frac{1}{-25.0} \] \[ \frac{1}{v} = -\frac{1}{15} + \frac{1}{25} \] Taking LCM of 15 and 25, which is 75:
\[ \frac{1}{v} = \frac{-5 + 3}{75} \] \[ \frac{1}{v} = \frac{-2}{75} \] \[ v = -\frac{75}{2} = -37.5\text{ cm} \] The negative sign indicates the image is formed in front of the mirror, so it is a real image.
Therefore, the screen should be placed \(37.5\text{ cm}\) in front of the mirror.
Now, finding magnification \(m\):
\[ m = -\frac{v}{u} = -\frac{-37.5}{-25.0} = -1.5 \] The negative magnification confirms the image is inverted.
Finding the size of the image \(h'\):
\[ m = \frac{h'}{h} \] \[ h' = m \times h = -1.5 \times 4.0\text{ cm} \] \[ h' = -6.0\text{ cm} \] The size of the image is \(6.0\text{ cm}\) (the negative sign indicates it is inverted).

Step 3: Final Answer:
Screen distance is \(37.5\text{ cm}\), nature is real and inverted, and size is \(6.0\text{ cm}\).