Question

A small candle, $2.5\text{cm}$ in size is placed at $27\text{cm}$ in front of a concave mirror of radius of curvature $36\text{cm}$. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

Answer 1

In the direction of incident ray, distance is considered positive.
From mirror formula:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Here,
$u=-27\text{cm}$
$f=\frac{R}{2}=-18\text{cm}$
Height of object $\left(h\right)=2.5\text{cm}$
Putting values, we get
$\frac{1}{v}+\left(\frac{1}{-27}\right)=\left(\frac{1}{-18}\right)$
The size of image is $5\text{cm}$.
If candle move closer to mirror
As $u\to f,v\to \infty );$
For u < f, image is virtual.
Final Answer: $v=-54\text{cm}$. The image is real, inverted and magnified. The size of the image is $5.0\text{cm}$. As $u\to f,v\to \infty$; for u < f, image is virtual.