Q114 / 120 · VITEEE 2015 · Physics
Two convex lenses of power $2 \mathrm{D}$ and $5 \mathrm{D}$ are separated by a distance $\frac{1}{3} \mathrm{~m}$. The power of optical system formed is
- A$+2 \mathrm{D}$
- B$-2 \mathrm{D}$
- C$-3 \mathrm{D}$
- Correct answerD$+3 \mathrm{D}$
Solution
Given:
$\begin{array}{l}
\mathrm{P}_{1}=2 \mathrm{D} ; \mathrm{P}_{2}=3 \mathrm{D} \\
\mathrm{d}=\frac{1}{3} \mathrm{~m}
\end{array}$
We know that $\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_{1}}+\frac{1}{\mathrm{f}_{2}}-\frac{\mathrm{d}}{\mathrm{f}_{1} \cdot \mathrm{f}_{2}}$ Equivatent power,
$\begin{aligned}
\therefore \quad & P=P_{1}+P_{2}-d P_{1} \cdot P_{2} \\
&=2+3-\frac{1}{3} \times 2 \times 3=3 D
\end{aligned}$
Hence, the correct answer is option: 'D'.