Answer to Question #170571 in Optics for Innocent

Question

An erect image,four times the size of the object is obtained with a concave mirror of radius of curvature 48cm. What is the position of the objects

Answer 1

Let's write the magnification equation:

$M=\dfrac{h_i}{h_o}=\dfrac{-d_i}{d_o}=4,$ $d_i=-4d_o.$

Let's wtrite the equation of thin lens:

$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f},$ $\dfrac{1}{d_o}+\dfrac{1}{-4d_o}=\dfrac{2}{R},$ $d_o=\dfrac{3}{8}R=\dfrac{3}{8}\cdot48\ cm=18\ cm.$

Finally, we can find the image distance:

$d_i=-4d_o=-4\cdot18\ cm=-72\ cm.$

The sign minus means that the image is virtual