MHT CET · Physics · Ray Optics
There are four convex lenses \(\mathrm{L}_{1}, \mathrm{~L}_{2}, \mathrm{~L}_{3}\) and \(\mathrm{L}_{4}\) of focal length \(2,4,6\) and \(8 \mathrm{~cm}\) respectively. Two of these lenses form a telescope of length \(10 \mathrm{~cm}\) and magnifying power 4 . The objective and eye lenses are respectively
- A L1, L2
- B L1, L4
- C L2, L3
- D L4, L1
Answer & Solution
Correct Answer
(D) L4, L1
Step-by-step Solution
Detailed explanation
(C)
\(\begin{array}{l}
\mathrm{f}_{0}+\mathrm{f}_{\mathrm{e}}=10 \\
\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}=4 \quad \therefore \mathrm{f}_{0}=4 \mathrm{f}_{\mathrm{e}} \\
5 \mathrm{f}_{\mathrm{e}}=10 \quad \therefore \mathrm{f}_{\mathrm{e}}=2 \quad \mathrm{f}_{0}=8
\end{array}\)
So, \(\mathrm{L}_{1}\) and \(\mathrm{L}_{4}\) were used
\(\begin{array}{l}
\mathrm{f}_{0}+\mathrm{f}_{\mathrm{e}}=10 \\
\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}=4 \quad \therefore \mathrm{f}_{0}=4 \mathrm{f}_{\mathrm{e}} \\
5 \mathrm{f}_{\mathrm{e}}=10 \quad \therefore \mathrm{f}_{\mathrm{e}}=2 \quad \mathrm{f}_{0}=8
\end{array}\)
So, \(\mathrm{L}_{1}\) and \(\mathrm{L}_{4}\) were used
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