• # question_answer Diameter of a plano-convex lens is 6 cm and its thickness is 4mm. What is the focal length of the lens, if refractive index of the glass is 1.5?

According to the lens maker formula, $\frac{1}{f}=(\mu -1)\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]$ For plano-convex lens, ${{R}_{1}}=R$ and ${{R}_{2}}=\infty ,$ $\mu =1.5$ (given) $\therefore$ $\frac{1}{f}=(\mu -1)\left[ \frac{1}{R}-\frac{1}{\infty } \right]\Rightarrow f=\frac{R}{(\mu -1)}$ $f=2R\,\,\left( \text{putting},\mu =\frac{3}{2} \right)$                              ?(i) It r is the radius and y is the thickness of the lens (at the centre), we have from the figure,             ${{R}^{2}}={{r}^{2}}+{{(R-Y)}^{2}}$ $\therefore$      $R=\frac{{{r}^{2}}+{{y}^{2}}}{2y}$                         Since, $r=\frac{6}{2}=3cm,$                         y = 4 mm $\therefore$      $R\simeq \frac{{{r}^{2}}}{2y}$                       $({{r}^{2}}+{{y}^{2}}\approx {{r}^{2}},\because r>y)$             $=\frac{9}{2\times 4\times {{10}^{-1}}}=1125\,cm$ Putting the value of fi into Eq. (i), we get $f=2\times R=2\times 11.25$      $(\therefore f=22.5\,cm)$