Q115 / 150 · MHT CET 2021 (23 Sep Shift 1) · Physics

The radii of curvature of both the surface of a convex lens of focal length ' $\mathrm{f}$ ' and focal power ' $\mathrm{P}$ ' are equal. One of the surfaces is made plane by grinding. The new focal length and focal power of the lens is respectively

  • A
    $\frac{1}{2} \mathrm{f}, 2 \mathrm{P}$
  • Correct answer
    B
    $2 \mathrm{f}, \frac{\mathrm{P}}{2}$
  • C
    $\sqrt{\frac{2}{f}}, \sqrt{\frac{P}{2}}$
  • D
    $\frac{2 \mathrm{f}}{3}, \frac{2}{3} \mathrm{P}$

Solution

$\begin{aligned} & \mathrm{P}=\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}\right)=(\mu-1) \frac{2}{\mathrm{R}} \\ & \mathrm{P}^{\prime}=\frac{1}{\mathrm{f}^{\prime}}=(\mu-1) \frac{1}{\mathrm{R}} \\ & \therefore \mathrm{P}^{\prime}=\frac{\mathrm{P}}{2} \text { and } \mathrm{f}^{\prime}=2 \mathrm{f}\end{aligned}$

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

Relevant Questions