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A thin lens made of glass of refractive index \mu =1.5 has a focal length equal to \mathrm{12\ cm} in air. It is now immersed in water \mathrm{\left ( \mu =\frac{4}{3} \right )} . Its new focal length is:

Option: 1

\mathrm{48\ cm}


Option: 2

\mathrm{36\ cm}


Option: 3

\mathrm{24\ cm}


Option: 4

\mathrm{12\ cm}


Answers (1)

best_answer

Focal length in air is given by, \frac{1}{\mathrm{f}_{\mathrm{a}}}=\left(\mathrm{a} \mu_{\mathrm{g}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)
The focal length of lens immersed in water is given by, \frac{1}{\mathrm{f}_l}=\left(l \mu_{\mathrm{g}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)
where, \mathrm{R_{1},R_{2}} are radii of curvatures of the two surfaces of lens and \mathrm{_{l}\mu _{g}} is refractive index of glass with respect to liquid.
\mathrm{Also, l \mu_{\mathrm{g}}=\frac{\mathrm{a}_{\mathrm{a}} \mu_{\mathrm{a}}}{\mu_l}} \\ \mathrm{Given, { }_{\mathrm{a}} \mu_{\mathrm{g}}=1.5, \mathrm{f}_{\mathrm{a}}=12 \mathrm{~cm}\ and\ { }_{\mathrm{a}} \mu_l=4 / 3}
\begin{aligned} & \therefore \frac{\mathrm{f}_l}{\mathrm{f}_{\mathrm{a}}}=\frac{\left({ }_{\mathrm{a}} \mu_{\mathrm{g}}-1\right)}{\left({ }_l \mu_{\mathrm{g}}-1\right)} \\ & \frac{\mathrm{f}_1}{12}=\frac{(1.5-1)}{\left(\frac{1.5}{4 / 3}-1\right)}=\frac{0.5 \times 4}{0.5} \\ & \Rightarrow \mathrm{f}_l=4 \times 12=48 \mathrm{~cm} \end{aligned}

Posted by

Gautam harsolia

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