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

Option: 1

\mathrm{f}


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

Let R be the radius of curvature of each surface.
\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}\right) \quad \therefore \quad \mathrm{R}=\mathrm{f}

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{\text { Or } \frac{1}{f^{\prime}}=-\frac{2}{3 f}}

\mathrm{ \text { Using, } \frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2}+\frac{1}{\mathrm{f}_3^{\prime}} \\ }

\mathrm{ \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}} \\ }

\mathrm{ \therefore \quad \mathrm{F}=\frac{3 \mathrm{f}}{4} }

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