Question

A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm, then

• A. at a distance of 54 cm from the mirror should a screen be placed in order to obtain a sharp image
  
• B. image is inverted and real
• C. If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror
  
• D. at a distance of 60 cm from the mirror should a screen be placed in order to obtain a sharp image
  

Answer 1

Answer: (A), (B), (C)

If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror


Size of the candle, $h=2.5cm$
Image size $=h^{\prime }$
Object distance, $u=–27cm$
Radius of curvature of the concave mirror, $R=–36cm$
Focal length of the concave mirror,
$f=\frac{R}{2}=-18cm$
$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
$=\frac{1}{-18}-\frac{1}{-27}=\frac{-3+2}{54}=\frac{-1}{54}$
$\therefore v=-54cm$
Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.
The magnification of the image is given as:
$m=\frac{h^{{}^{\prime }}}{h}=\frac{-v}{u}$
$\therefore h^{{}^{\prime }}=\frac{-v}{u}\times h$
$-\left(\frac{-54}{-27}\right)\times 2.5=-5cm$
The height of the candle’s image is $5cm$. The negative sign indicated that the image is inverted and real.
If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.