An equiconvex lens forms a two times enlarged real image when an object is kept 16 cm from it. The lens is cut into two identical plano-convex…

Finding the position of the image Nature $\displaystyle \frac{1}{f}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$ $\displaystyle \frac{1}{f}=\frac{1}{32}+\frac{1}{16}=\frac{1+2}{32}=\frac{3}{32}$ $\displaystyle f=\frac{32}{3} c m$ When lens is cut into two equal halves the new focal length $\displaystyle =f^{\prime}=2 f=\frac{64}{3} \mathrm{~cm}$ $\displaystyle \frac{1}{f^{\prime}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \Rightarrow \frac{3}{64}=\frac{1}{v}+\frac{1}{16}$ $\displaystyle \mathrm{v}=-64 \mathrm{~cm}$ Image formed is virtual \