Physics - Geometrical Optics Question with Solution | TestHub

The student should use a lens which forms image at distance of 15 cm of the object placed at 3 m i.e., object distance $u=-3\mathrm{m}=-300\mathrm{cm}$, image distance $v=-15\mathrm{cm}$ From lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ we get $\frac{1}{-15}-\frac{1}{-300}=\frac{1}{f}$ or $\frac{1}{f}=\frac{1}{300}-\frac{1}{15}=-\frac{19}{300}$ or $f=\frac{-300}{19}=-15.8\mathrm{cm}$ (concave lens) Now power of the lens is $P=\frac{100}{f(\mathrm{cm})}=\frac{100}{-15.8}=-6.33\mathrm{D}$