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The refractive index of a material of a planoconcave lens is $5 / 3$, the radius of curvature is $0.3 \mathrm{~m}$. The focal length of the lens in air is
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Verified Answer
The correct answer is:
$-0.45 \mathrm{~m}$
Lens maker's formula
$$
\frac{1}{f}=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]
$$
where, $R_2=\infty, R_1=0.3 \mathrm{~m}$
$$
\begin{array}{llrl}
\therefore & & \frac{1}{f} & =\left(\frac{5}{3}-1\right)\left(\frac{1}{0.3}-\frac{1}{\infty}\right) \\
& \Rightarrow & \frac{1}{f} & =\frac{2}{3} \times \frac{1}{0.3} \\
\text { or } & & f & =0.45 \mathrm{~m}
\end{array}
$$
$$
\frac{1}{f}=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]
$$
where, $R_2=\infty, R_1=0.3 \mathrm{~m}$
$$
\begin{array}{llrl}
\therefore & & \frac{1}{f} & =\left(\frac{5}{3}-1\right)\left(\frac{1}{0.3}-\frac{1}{\infty}\right) \\
& \Rightarrow & \frac{1}{f} & =\frac{2}{3} \times \frac{1}{0.3} \\
\text { or } & & f & =0.45 \mathrm{~m}
\end{array}
$$
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