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The focal length of the lens of refractive index \mathrm{\left ( \mu =1.5 \right )} in air is \mathrm{10\ cm}. If air is replaced by water of \mathrm{\mu=\frac{4}{3}}, its focal length is:

Option: 1

\mathrm{20\ cm}


Option: 2

\mathrm{30\ cm}


Option: 3

\mathrm{40\ cm}


Option: 4

\mathrm{25\ cm}


Answers (1)

best_answer

According to lens maker’s formula, The focal length of the lens in air is
\mathrm{\begin{aligned} & \frac{1}{\mathrm{f}_{\text {air }}}=\left(\frac{\mu_l}{\mu_{\mathrm{a}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)=\left(\frac{3 / 2}{1}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\ & \frac{1}{\mathrm{f}_{\text {air }}}=\frac{1}{2}\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right).............(i) \end{aligned}}
The focal length of the lens in water is
\mathrm{\begin{aligned} & \frac{1}{\mathrm{f}_{\text {water }}}=\left(\frac{\mu_1}{\mu_{\mathrm{w}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)=\left(\frac{3 / 2}{4 / 3}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\ & \frac{1}{\mathrm{f}_{\text {water }}}=\frac{1}{8}\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) ............(ii)\end{aligned}}
Divide \mathrm{(i)} by \mathrm{(ii)}, we get
\mathrm{\begin{aligned} & \mathrm{\frac{f_{\text {water }}}{f_{\text {air }}}=4} \\ & \mathrm{f_{\text {water }}=4 \times f_{\text {air }}=4 \times 10 \mathrm{~cm}=40 \mathrm{~cm}} \end{aligned}}

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Rishi

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