Question 178 of 295

The graph between $\dfrac{1}{u}$ and $\dfrac{1}{v}$ for a thin convex lens in order to determine its focal length is plotted as shown in the figure. The refractive index of lens is $1.5$ and its both the surfaces have same radius of curvatures $R$. The value of $R$ will be _____ $cm$.
(Where $u=$ object distance, $v=$ image distance)

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Answer: 10

For point $B,\dfrac{1}{u}=-0.10{cm}^{-1},\dfrac{1}{v}=0$
$∴$ Thus, $u=-10cm,v=\infty$
Using lens formula,
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=0-\left(-10\right)$
$\Rightarrow f=10cm$
Now, using Lens maker's formula,
$\dfrac{1}{f}=\left(\mu -1\right)\left(\dfrac{1}{{R}_{1}}-\dfrac{1}{{R}_{2}}\right)$, where, $\mu$ is refractive index of lens.
$\Rightarrow \dfrac{1}{10cm}=\left(1.5-1\right)\left(\dfrac{2}{R}\right)=\dfrac{1}{R} \Rightarrow R=10cm$