Q141 / 160 · TS EAMCET 2024 (09 May Shift 1) · Physics
For a combination of two convex lenses of focal lengths ' $f_1{ }^{\prime}$ and ' $f_2$ ' to act as a glass slab, the distance of separation between them is
- Correct answerA$f_1+f_2$
- B$f_1 \sim f_2$
- C$\frac{f_1+f_2}{2}$
- D$\frac{f_1 \sim f_2}{2}$
Solution
In case of glass slab $F=\infty$
If equivalent focal length of convex lenses of focal lengths $f_1$ and $f_2$ separated by a distance $d$ (suppose) is infinity then the combination of lenses will act as a glass slab
$\begin{aligned}
& \frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2} \\
& \text { or, } \frac{1}{\infty}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2} \Rightarrow 0=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2} \\
& \text { or, } 0=\frac{f_1+f_2}{f_1+f_2}-\frac{d}{f_1 f_2} \therefore d=f_1+f_2
\end{aligned}$
Hence, the correct answer is option: 'A'.