Answer to Question #115342 in Optics for John

Question

When an object is placed 20cm in front of a spherical mirror produces an erect image which is 1/4 of the size of the object. Determine the type of the mirror and radius of curvature of the mirror

Answer 1

The mirror is convex (from general theory).

So, we have $\frac{R-d_i}{R+d_o}=\frac{1}{4} \rightsquigarrow (3R-d_o)/4=d_i$

And $-\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}=\frac{2}{R} \rightsquigarrow (d_o-d_i)R=2d_od_i$

Hence, $(5d_o-3R)R=2d_o(3R-d_o) \rightsquigarrow 3R^2+d_oR-2d_o^2=0$

For $d_o=20\colon R=40/3 \approx 13.33\,cm$