A symmetric glass lens of index 1.50 has a focal length of -17.5 cm in air. A plastic bag having pressurized air, forming an "air lens" to be used underwater. Find its focal length. Additional Materials eBook
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The refractive index of the glass lens, $n_g = 1.50$. The focal length of the glass lens in air, $f_{air,glass} = -17.5$ cm. The refractive index of air, $n_{air} = 1.00$. Show more…
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A thin glass lens $(n=1.50)$ has a focal length of $+10 \mathrm{~cm}$ in air. Compute its focal length in water $(n=1.33)$. Using $$\frac{1}{f}=\left(\frac{n_{1}}{n_{2}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$$ we get $$\begin{array}{ll} \text { For air: } & \frac{1}{10}=(1.50-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\ \text { For water: } & \frac{1}{f}=\left(\frac{1.50}{1.33}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \end{array}$$ Divide one equation by the other to obtain $f=5.0 / 0.128=39 \mathrm{~cm}$.
Determine the focal length in air of a thin spherical planar-convex lens having a radius of curvature of $50.0 \mathrm{mm}$ and an index of 1.50. What, if anything, would happen to the focal length if the lens were placed in a tank of water?
A thin glass lens $(n=1.50)$ has a focal length of $+10 \mathrm{~cm}$ in air. Compute its focal length in water $(n=1.33)$. Using $$ \frac{1}{f}=\left(\frac{n_{1}}{n_{2}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) $$ we get For air: $\quad \frac{1}{10}=(1.50-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$ $$ \text { \mathrm{\{} F o r ~ w a t e r : ~ } \quad \frac{1}{f}=\left(\frac{1.50}{1.33}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) $$ Divide one equation by the other to obtain $f=5.0 / 0.128=39 \mathrm{~cm}$.
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