Explanation:
A: Given,
\(\mathrm{R}_1=\mathrm{R}_2=\mathrm{R}\)
Using lens maker formula,
\(\begin{aligned}
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{(-\mathrm{R})}\right) \\
\frac{1}{\mathrm{f}}=\frac{2(\mu-1)}{\mathrm{R}}
\end{aligned}\)
Power of lens, \(P=\frac{1}{f}=\frac{2(\mu-1)}{R}=4 D\)
\(\frac{(\mu-1)}{R}=2 \mathrm{D}\)
Now, the lens is cut perpendicular to its axis-
\(\mathrm{R}_1=\mathrm{R}, \& \mathrm{R}_2=\infty\)
Again -
\(\begin{aligned}
\frac{1}{\mathrm{f}^{\prime}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
\frac{1}{\mathrm{f}^{\prime}}=(\mu-1)\left(\frac{1}{\mathrm{R}}-\frac{1}{\infty}\right)=\frac{(\mu-1)}{\mathrm{R}}
\end{aligned}\)
Power of lens, \(\mathrm{P}^{\prime}=\frac{1}{\mathrm{f}^{\prime}}=\frac{(\mu-1)}{\mathrm{R}}=2 \mathrm{D} \quad\{\) from (i) \(\}\)
\(\mathrm{P}^{\prime}=2 \mathrm{D}\)