Q92P - University Physics with Modern Physics

Question

(a) Repeat the derivation of Eq. (34.19) for the case in which the lens is totally immersed in a liquid of refractive index ${\mathit{n}}_{\mathbf{l}\mathbf{i}\mathbf{q}}$ . (b) A lens is made of glass that has refractive index 1.60. In air, the lens has focal length +18.0 cm. What is the focal length of this lens if it is totally immersed in a liquid that has refractive index 1.42?

Step-by-Step Solution

Answer

1. The derivated equation is $\frac{1}{\mathit{f}}=\left(\frac{\mathit{n}}{{\mathit{n}}_{\mathbf{l}\mathbf{i}\mathbf{q}}}-1\right)\left(\frac{1}{{\mathit{R}}_1}-\frac{1}{{\mathit{R}}_2}\right)$
2. The focal length of the lens when it is completely immersed in a liquid with refractive index 1.42 is $\mathbf{85}\mathbf{.}\mathbf{2}\mathit{c}\mathit{m}$ 

Step 1: Lensmaker’s equation for single surface and two surface.

$\frac{{\mathit{n}}_{\mathbf{a}}}{\mathit{s}}+\frac{{\mathit{n}}_{\mathbf{b}}}{\mathit{s}\mathbf{\text{'}}}=\frac{{\mathit{n}}_{\mathbf{b}}\mathbf{-}{\mathit{n}}_{\mathbf{a}}}{\mathit{R}}$ where n is the refraction index and R is the curvatures of the lens.

$\frac{1}{\mathit{f}}=\left(\mathit{n}-1\right)\mathbf{(}\frac{1}{{\mathit{R}}_1}-\frac{1}{{\mathit{R}}_2}\mathbf{)}$ where n is the refraction index and R is the curvatures of the lens.

Step 2:Derivation for the lens that is immersed in a liquid.

Using the lensmaker’s formula for both the surfaces,

$$\begin{gathered} \frac{{\mathbf{n}}_{\mathbf{a}}}{{\mathbf{s}}_{\mathbf{1}}}\mathbf{+}\frac{{\mathbf{n}}_{\mathbf{b}}}{\mathbf{s}{\mathbf{\text{'}}}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{n}}_{\mathbf{b}}\mathbf{-}{\mathbf{n}}_{\mathbf{a}}}{{\mathbf{R}}_{\mathbf{1}}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathit{i} \\ \frac{{\mathbf{n}}_{\mathbf{b}}}{{\mathbf{s}}_{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{n}}_{\mathbf{c}}}{\mathbf{s}{\mathbf{\text{'}}}_{\mathbf{2}}}\mathbf{=}\frac{{\mathbf{n}}_{\mathbf{c}}\mathbf{-}{\mathbf{n}}_{\mathbf{b}}}{{\mathbf{R}}_{\mathbf{2}}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathit{i}\mathit{i} \end{gathered}$$ 

Here  ${\mathit{n}}_{\mathbf{a}}\mathbf{ }\mathbf{\&}\mathbf{ }{\mathit{n}}_{\mathbf{c}}$ are liquid and ${\mathit{n}}_{\mathbf{b}}$ is air and ${\mathit{s}}_2$  equals $-\mathit{s}{\mathbf{\text{'}}}_2$.

Adding the equations i and ii.

$$\begin{gathered} \frac{{\mathbf{n}}_{\mathbf{l}\mathbf{a}\mathbf{q}}}{{\mathbf{s}}_{\mathbf{1}}}\mathbf{+}\frac{{\mathbf{n}}_{\mathbf{l}\mathbf{i}\mathbf{q}}}{\mathbf{s}{\mathbf{\text{'}}}_{\mathbf{2}}}=\left(n-n_{liq}\right)\left(\frac{1}{{\mathit{R}}_1}-\frac{1}{{\mathit{R}}_2}\right) \\ \frac{1}{{\mathbf{s}}_{\mathbf{1}}}\mathbf{+}\frac{1}{\mathbf{s}{\mathbf{\text{'}}}_{\mathbf{2}}}=\mathbf{(}\frac{\mathbf{n}}{{\mathbf{n}}_{\mathbf{liq}}}\mathbf{-}\mathbf{1}\mathbf{)}\left(\frac{1}{{\mathit{R}}_1}-\frac{1}{{\mathit{R}}_2}\right) \\ \frac{1}{\mathit{f}}=\left(\frac{\mathit{n}}{{\mathbf{n}}_{\mathbf{liq}}}-1\right)\left(\frac{1}{{\mathit{R}}_1}-\frac{1}{{\mathit{R}}_2}\right) \end{gathered}$$ 

Hence proved.

Step 3:Calculating the required focal length.

Equation for air, $\frac{1}{{\mathit{f}}_{\mathbf{a}\mathbf{i}\mathbf{r}}}=\left(\mathit{n}-1\right)(\frac{1}{{\mathit{R}}_{\mathbf{1}}}-\frac{1}{{\mathit{R}}_{\mathbf{2}}})$ 

Equation for liquid derived in part a),  $\frac{\mathbf{1}}{{\mathbf{f}}_{\mathbf{a}\mathbf{i}\mathbf{r}}}\mathbf{=}\left(\frac{n}{n_{liq}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Dividing both the equations,

$$\begin{gathered} {\mathit{f}}_{\mathbf{l}\mathbf{i}\mathbf{q}}\mathbf{=}{\mathit{f}}_{\mathbf{a}\mathbf{i}\mathbf{r}}\frac{\left(n-1\right)}{\left({\displaystyle \frac{\mathbf{n}}{{\mathbf{n}}_{\mathbf{l}\mathbf{i}\mathbf{q}}}}-1\right)} \\ =\left(18\mathbf{cm}\right)\frac{\left(1.6-1\right)}{\left([1.6-1.42]-1\right)} \\ =85.2\mathbf{cm} \end{gathered}$$