MHT CET · Physics · Ray Optics
Two thin lenses have a combined power of +9 D. When they are separated by a distance of 20 cm , their equivalent power becomes \(+\frac{27}{5} \mathrm{D}\). The power of both the lenses in dioptre are respectively
- A 4,5
- B 3,6
- C 2,7
- D 1,8
Answer & Solution
Correct Answer
(B) 3,6
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& P_1+P_2=9 D \ldots . \text { (i) } \\
& P_1+P_2-d_1 P_2=\frac{27}{5} \ldots . \text { (ii) }
\end{aligned}\)
where \(\mathrm{d}=20 \mathrm{~cm}=0.2 \mathrm{~m}\)
Substituting value of \(\left(\mathrm{P}_1+\mathrm{P}_2\right)\) in equation (ii) we get
\(\begin{array}{ll}
& 9-0.2 P_1 P_2=\frac{27}{5} \\
\therefore & 0.2 P_1 P_2=9-\frac{27}{5}=\frac{18}{5} \\
\therefore & P_1 P_2=\frac{18}{0.2 \times 5}=18
\end{array}\)
Since \(P_1+P_2=9\) and \(P_1 P_2=18\)
\(\therefore \quad \mathrm{P}_1=3\) and \(\mathrm{P}_2=6\)
& P_1+P_2=9 D \ldots . \text { (i) } \\
& P_1+P_2-d_1 P_2=\frac{27}{5} \ldots . \text { (ii) }
\end{aligned}\)
where \(\mathrm{d}=20 \mathrm{~cm}=0.2 \mathrm{~m}\)
Substituting value of \(\left(\mathrm{P}_1+\mathrm{P}_2\right)\) in equation (ii) we get
\(\begin{array}{ll}
& 9-0.2 P_1 P_2=\frac{27}{5} \\
\therefore & 0.2 P_1 P_2=9-\frac{27}{5}=\frac{18}{5} \\
\therefore & P_1 P_2=\frac{18}{0.2 \times 5}=18
\end{array}\)
Since \(P_1+P_2=9\) and \(P_1 P_2=18\)
\(\therefore \quad \mathrm{P}_1=3\) and \(\mathrm{P}_2=6\)
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