The radius of curvature of the face of planoconvex lens is $12 \mathrm{~cm}$ and its refractive index is 1.5. Then, the focal length of the lens is

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Aicharya · Accepted Answer

Given, refractive index of lens, $\mu=1.5$ Radius of curved face, $R_1=12 \mathrm{~cm}$ Radius of plane face, $R_2=\infty$

Now, using the expression of focal length, $ \frac{1}{f}=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right] $ Substituting the values, we get $ \begin{aligned} \frac{1}{f} & =(1.5-1)\left[\frac{1}{12}-\frac{1}{\infty}\right] \\ \frac{1}{f} & =0.5 \times \frac{1}{12} \\ \Rightarrow \quad f & =\frac{12}{0.5} \mathrm{~cm}=24 \mathrm{~cm} \end{aligned} $ Hence, the focal length of given plano-convex lens is $24 \mathrm{~cm}$

The radius of curvature of the face of planoconvex lens is $12 \mathrm{~cm}$ and its refractive index is 1.5…

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