Q144 / 180 · KCET 2011 · Physics

A planoconvex lens has a maximum thickness of $6 \mathrm{~cm}$. When placed on a horizontal table with the curved surface in contact with the table surface, the apparent depth of the bottommost point of the lens is found to be $4 \mathrm{~cm}$. If the lens is inverted such that the plane face of the lens is in contact with the surface of the table, the apparent depth of the centre of the plane face is found to be $\left(\frac{17}{4}\right) \mathrm{cm}$. The radius of curvature of the lens is

  • A
    $68 \mathrm{~cm}$
  • B
    $75 \mathrm{~cm}$
  • C
    $128 \mathrm{~cm}$
  • Correct answer
    D
    $34 \mathrm{~cm}$

Solution

$\begin{aligned} n &=\frac{\text { Real depth }}{\text { Apparent depth }} \\ n &=\frac{6}{4}=\frac{3}{2} \\ \frac{n_{1}}{u}-\frac{n_{2}}{v} &=\frac{n_{1} \sim n_{2}}{R} \\ \frac{1.5}{6}-\frac{4}{17} &=\frac{1.5-1}{R} \\ R &=34 \mathrm{~cm} \end{aligned}$

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

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