Chapter 27 - Optical Instruments - Problems and Conceptual Exercises - Page 972: 77

$1.8cm$

We know that $M_{total}=\frac{\theta^{\circ}}{\theta}$ $\implies M_{total}=\frac{15}{0.50}=30.0$ Now $f_{eyepiece}=\frac{f_{objective}}{M_{total}}$ We plug in the known values to obtain: $f_{eyepiece}=\frac{53}{30.0}=1.8cm$