Q144 / 160 · TS EAMCET 2021 (05 Aug Shift 2) · Physics

A point source is located at a distance of $20 cm$ from the front surface of a symmetrical glass biconvex lens with equal radii of curvature $5 cm .$ The distance at which image formed from the rear surface of this lens is [Given refractive index of the glass is $1.5$ ]

Solution

From lens makers formula for biconvex lens, we have

\frac{1}{f} = (μ - 1) \frac{2}{R}

\Rightarrow \frac{1}{f} = (1.5 - 1) \frac{2}{5} = \frac{1}{5}

From Lens equation, we have

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\Rightarrow \frac{1}{v} - \frac{1}{- 20} = \frac{1}{5}

\Rightarrow \frac{1}{v} = \frac{1}{5} - \frac{1}{20} = \frac{3}{20}

\Rightarrow v = \frac{20}{3} cm

Hence, the correct answer is option: 'A'.