Explanation:
B: Given, \(\mu=1.5\)
\(\mathrm{R}_2=100 \mathrm{~cm}, \mathrm{R}_1=\infty\)
Power of lens can be determined with the index of refraction and also the radius of curvature,
Power of lens \(=\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\)
Where \(R_1\) is radius of curvature of plane surface and \(R_2\) is concave surface,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{\infty}-\frac{1}{100}\right) \\
\frac{1}{\mathrm{f}}=\frac{-0.5}{100} \\
\frac{1}{\mathrm{f}}=-0.5 \mathrm{~m} \\
\mathrm{p}=\frac{1}{\mathrm{f}}=-0.5 \mathrm{D}
\end{aligned}\)
