JEE Main · 2004 · 24 Apr 2004 · Offline · Geometrical Optics

With its curved face silvered, the lens-mirror behaves as an equivalent mirror of power $\frac{1}{F}=\frac{2}{f_l}+\frac{1}{f_m}$. The lens power is $\frac{1}{f_l}=(n-1)\left(\frac{1}{\infty }-\frac{1}{-30}\right)=\frac{0.5}{30}=\frac{1}{60}$, so $f_l=60$ cm, while the silvered curved surface acts as a concave mirror of focal length $f_m=\frac{R}{2}=15$ cm. Then $\frac{1}{F}=\frac{2}{60}+\frac{1}{15}=\frac{1}{10}$, giving $F=10$ cm, and a real image equal in size to the object forms when the object stands at the centre of curvature, a distance $2F=20$ cm away.