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Two identical glass \mathrm{(\mu_{g}=\frac{3}{2})} equiconvex lenses of focal length \mathrm{f} are kept in contact. The space between the two lenses is filled with water \mathrm{(\mu _{\omega}=\frac{4}{3})} . The focal length of the combination is:

Option: 1

\mathrm{f}


Option: 2

\mathrm{\frac{f}{2}}


Option: 3

\mathrm{\frac{4f}{3}}


Option: 4

\mathrm{\frac{3f}{4}}


Answers (1)

best_answer

Let \mathrm{R} be the radius of curvature of each surface.

\mathrm{\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}\right) \quad \therefore \quad \mathrm{R}=\mathrm{f}} \\ \mathrm{For\ the\ water\ lens,\ \frac{1}{\mathrm{f}^{\prime}}=\left(\frac{4}{3}-1\right)\left(-\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}}\right)=\frac{1}{3}\left(-\frac{2}{\mathrm{f}}\right)}\\ \mathrm{or \quad \frac{1}{f^{\prime}}=-\frac{2}{3 f}}
\begin{aligned} & \text { Using, } \frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2}+\frac{1}{\mathrm{f}_3} \\ & \frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{f}^{\prime}}=\frac{2}{\mathrm{f}}-\frac{2}{3 \mathrm{f}}=\frac{4}{3 \mathrm{f}} \quad \therefore \quad \mathrm{F}=\frac{3 \mathrm{f}}{4} \end{aligned}

Posted by

Suraj Bhandari

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