Q142 / 160 · AP EAMCET 2023 (17 May Shift 2) · Physics

The radius of curvature of a convex lens is $40 \mathrm{~cm}$, for each surface. Its refractive index is 1.5 . Its focal length is

  • Correct answer
    A
    $40 \mathrm{~cm}$
  • B
    $20 \mathrm{~cm}$
  • C
    $80 \mathrm{~cm}$
  • D
    $30 \mathrm{~cm}$

Solution

From the formula $\begin{aligned} & \frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\ & \frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{2}{40}\right)=\frac{1}{40}\end{aligned}$ focal length, $f=40 \mathrm{~cm}$

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

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