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If the refractive index from air to glass is $\frac{3}{2}$ and that from air to water is $\frac{4}{3}$, then the ratio of focal lengths of a glass lens in water and in air is
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$4: 1$
Given, refractive index from air to glass,
${ }_{a} \mu_{g}=\frac{3}{2}$
Refractive index from air to water, ${ }_{a} \mu_{w}=\frac{4}{3}$
Using lens Maker's formula,
$\begin{aligned} \frac{1}{f} &=\left({ }_{1} \mu_{2}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\ \therefore \quad \frac{f_{\text {water }}}{f_{\text {air }}} &=\frac{\left({ }_{a} \mu_{g}-1\right)}{\left({ }_{w} \mu_{g}-1\right)}=\frac{\left({ }_{a} \mu_{g}-1\right)}{\left(\frac{{ }_{a} \mu_{g}}{{ }_{a} \mu_{w}}-1\right)} \\ &=\frac{\left(\frac{3}{2}-1\right)}{\left(\frac{3}{2}\right)}=\frac{1}{2} \times \frac{8}{1}=\frac{4}{1} \text { or } 4: 1 \end{aligned}$
${ }_{a} \mu_{g}=\frac{3}{2}$
Refractive index from air to water, ${ }_{a} \mu_{w}=\frac{4}{3}$
Using lens Maker's formula,
$\begin{aligned} \frac{1}{f} &=\left({ }_{1} \mu_{2}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\ \therefore \quad \frac{f_{\text {water }}}{f_{\text {air }}} &=\frac{\left({ }_{a} \mu_{g}-1\right)}{\left({ }_{w} \mu_{g}-1\right)}=\frac{\left({ }_{a} \mu_{g}-1\right)}{\left(\frac{{ }_{a} \mu_{g}}{{ }_{a} \mu_{w}}-1\right)} \\ &=\frac{\left(\frac{3}{2}-1\right)}{\left(\frac{3}{2}\right)}=\frac{1}{2} \times \frac{8}{1}=\frac{4}{1} \text { or } 4: 1 \end{aligned}$
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