ExamBro
ExamBro
MHT CET · Physics · Ray Optics

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

  1. A \(\frac{1}{2} \mathrm{f}, 2 \mathrm{P}\)
  2. B \(2 \mathrm{f}, \frac{\mathrm{P}}{2}\)
  3. C \(\sqrt{\frac{2}{f}}, \sqrt{\frac{P}{2}}\)
  4. D \(\frac{2 \mathrm{f}}{3}, \frac{2}{3} \mathrm{P}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(2 \mathrm{f}, \frac{\mathrm{P}}{2}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \mathrm{P}=\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}\right)=(\mu-1) \frac{2}{\mathrm{R}} \\ & \mathrm{P}^{\prime}=\frac{1}{\mathrm{f}^{\prime}}=(\mu-1) \frac{1}{\mathrm{R}} \\ & \therefore \mathrm{P}^{\prime}=\frac{\mathrm{P}}{2} \text { and } \mathrm{f}^{\prime}=2 \mathrm{f}\end{aligned}\)
Same subject
Explore more questions on app