• # question_answer A glass hemisphere of radius R and material having refractive index 1.5 is silvered on its flat face as shown in the figure. A small object of height h is located at a distance 2R from the surface of hemisphere. The final image will be formed A)   at a distance of R from silvered surface, on right side.B)  on the object itself.C)  at the hemispherical surface.D)  at a distance of 2R from the silvered surface on left side.

Here, three optical phenomena take place-first refraction, then reflection and finally refraction. $\frac{1.5}{v}-\frac{1}{-2R}=\frac{1.5-1}{R}$ $\Rightarrow$$\frac{1.5}{v}=0\Rightarrow v=\infty$ This means that, rays after refraction get parallel to axis and strike the mirror normally, get retraced back and the final image is formed at the same place, where the object is and of the same size. Image would be real. Hence, the correction option is [b].