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