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A convex lens of focal length 25 cm and made of glass
with refractive index 1.5 is immersed in water. The
absolute change in focal length of the glass is
[Use refractive index of water = $\frac{4}{3}$ ]
Options:
with refractive index 1.5 is immersed in water. The
absolute change in focal length of the glass is
[Use refractive index of water = $\frac{4}{3}$ ]
Solution:
1091 Upvotes
Verified Answer
The correct answer is:
75 cm
We have
$$
\begin{aligned}
& \frac{1}{\mathrm{f}_{\text {water }}}=\left({ }^{\mathrm{w}} \mu_{\mathrm{g}}-1\right)\left(\frac{2}{\mathrm{R}}\right) \\
& \text { and, } \frac{1}{\mathrm{f}_{\mathrm{a}}}=\left({ }^{\mathrm{a}} \mu_{\mathrm{g}}-1\right)\left(\frac{2}{\mathrm{R}}\right) \\
& \text { So, } \frac{\mathrm{f}_{\mathrm{a}}}{\mathrm{f}_{\mathrm{w}}}=\frac{\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}-1}{\mu_{\mathrm{g}}-1}=\frac{\frac{1}{8}}{\frac{1}{2}}=\frac{1}{4} \\
& \Rightarrow \mathrm{f}_{\mathrm{w}}=4 \mathrm{f}_{\mathrm{a}}=100 \mathrm{~cm} \\
& \text { Now, } \Delta \mathrm{f}=(100-25) \mathrm{cm}=75 \mathrm{~cm}
\end{aligned}
$$
$$
\begin{aligned}
& \frac{1}{\mathrm{f}_{\text {water }}}=\left({ }^{\mathrm{w}} \mu_{\mathrm{g}}-1\right)\left(\frac{2}{\mathrm{R}}\right) \\
& \text { and, } \frac{1}{\mathrm{f}_{\mathrm{a}}}=\left({ }^{\mathrm{a}} \mu_{\mathrm{g}}-1\right)\left(\frac{2}{\mathrm{R}}\right) \\
& \text { So, } \frac{\mathrm{f}_{\mathrm{a}}}{\mathrm{f}_{\mathrm{w}}}=\frac{\frac{\mu_{\mathrm{g}}}{\mu_{\mathrm{w}}}-1}{\mu_{\mathrm{g}}-1}=\frac{\frac{1}{8}}{\frac{1}{2}}=\frac{1}{4} \\
& \Rightarrow \mathrm{f}_{\mathrm{w}}=4 \mathrm{f}_{\mathrm{a}}=100 \mathrm{~cm} \\
& \text { Now, } \Delta \mathrm{f}=(100-25) \mathrm{cm}=75 \mathrm{~cm}
\end{aligned}
$$
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