MHT CET · Physics · Ray Optics
A double convex lens of focal length ' \(F\) ' is cut into two equal parts along the vertical axis. The focal length of each part will be
- A \(2 \mathrm{~F}\)
- B F
- C \(\frac{\mathrm{F}}{2}\)
- D \(4 \mathrm{~F}\)
Answer & Solution
Correct Answer
(A) \(2 \mathrm{~F}\)
Step-by-step Solution
Detailed explanation
For bifocal convex lens
\(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\)
\(=\frac{(\mu-1) \times 2}{R} \quad \ldots .\left(R_1=R_2=R\right)\)
For plane surface \(\mathrm{R}_2=\infty\)
For half plane-convex lens
\(\begin{aligned}
& \frac{1}{\mathrm{f}^{\prime}}=(\mu-1) \frac{1}{\mathrm{R}} \\
& \frac{1 / \mathrm{f}}{1 / \mathrm{f}^{\prime}}=\frac{(\mu-1)}{\mathrm{R}} \times 2 \times \frac{\mathrm{R}}{\mu-1}=2 \\
& \frac{\mathrm{f}^{\prime}}{\mathrm{f}}=2 \\
& \mathrm{f}^{\prime}=2 \mathrm{f}
\end{aligned}\)
As focal length is \(F, f^{\prime}=2 F\).
\(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\)
\(=\frac{(\mu-1) \times 2}{R} \quad \ldots .\left(R_1=R_2=R\right)\)
For plane surface \(\mathrm{R}_2=\infty\)
For half plane-convex lens
\(\begin{aligned}
& \frac{1}{\mathrm{f}^{\prime}}=(\mu-1) \frac{1}{\mathrm{R}} \\
& \frac{1 / \mathrm{f}}{1 / \mathrm{f}^{\prime}}=\frac{(\mu-1)}{\mathrm{R}} \times 2 \times \frac{\mathrm{R}}{\mu-1}=2 \\
& \frac{\mathrm{f}^{\prime}}{\mathrm{f}}=2 \\
& \mathrm{f}^{\prime}=2 \mathrm{f}
\end{aligned}\)
As focal length is \(F, f^{\prime}=2 F\).
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