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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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Verified Answer
The correct answer is:
$24 \mathrm{~cm}$
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 $24 \mathrm{~cm}$.
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 $24 \mathrm{~cm}$.
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