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
- A \(\frac{1}{2} \mathrm{f}, 2 \mathrm{P}\)
- B \(2 \mathrm{f}, \frac{\mathrm{P}}{2}\)
- C \(\sqrt{\frac{2}{f}}, \sqrt{\frac{P}{2}}\)
- D \(\frac{2 \mathrm{f}}{3}, \frac{2}{3} \mathrm{P}\)
Answer & Solution
Correct Answer
(B) \(2 \mathrm{f}, \frac{\mathrm{P}}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{P}=\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}\right)=(\mu-1) \frac{2}{\mathrm{R}} \\ & \mathrm{P}^{\prime}=\frac{1}{\mathrm{f}^{\prime}}=(\mu-1) \frac{1}{\mathrm{R}} \\ & \therefore \mathrm{P}^{\prime}=\frac{\mathrm{P}}{2} \text { and } \mathrm{f}^{\prime}=2 \mathrm{f}\end{aligned}\)
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