Chapter 32, Problem 98P

(a)

To determine

The focal length of a thin lens in water is given by $f_w=f_a\left(\raisebox{1ex}{$n_w$}\!\left/ \!\raisebox{-1ex}{$n_a$}\right.\right)\left(\raisebox{1ex}{$\left(n-n_a\right)$}\!\left/ \!\raisebox{-1ex}{$\left(n-n_w\right)$}\right.\right)$ .

Explanation of Solution

Given:

A thin lens whose focal length in air is $f_a$ .

Introduction:

The focal length of a lens changes with change in the medium it is present.

The equation for refraction at a single surface relates the object distance, image distance, refractive indices of the media and the radius of curvature of the single surface lens.

Write the expression for the equation for refraction at water-lens interface.

  $\frac{n_w}{u}+\frac{n}{v\text{'}}=\frac{n-n_w}{R_1}$ …… (1)

Here, $n_w$ is the refractive index of water, $n$ is the refractive index of air, $R_1$ is the radius of curvature, $v\text{'}$ is the image distance and $u$ is object distance.

Write the expression for the equation for refraction at lens-water interface.

  $\frac{n}{-v\text{'}}+\frac{n_w}{v}=\frac{n_w-n}{R_2}$ …… (2)

Here, $n_w$ is the refractive index of water, $n$ is the refractive index of air, $R_2$ is the radius of curvature, $v$ is the image distance and $v\text{'}$ is object distance.

Add equation (1) and (2).

  $n_w\left(\frac{1}{u}+\frac{1}{v}\right)=\left(n-n_w\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ …… (3)

Write the expression for thin lens equation.

  $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$ …… (4)

Here, $f$ is the focal length of lens, $v$ is the image distance and $u$ is the object distance.

Substitute $\frac{1}{f_w}$ for $\frac{1}{u}+\frac{1}{v}$ in equation (3).

  $\frac{n_w}{f_w}=\left(n-n_w\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ …… (5)

Here, $f_w$ is the focal length of lens in water.

The focal length of a lens depends on the refractive index of lens, the medium in which lens is present and the radii of curvature of lens. It is expressed in form of lens maker’s formula.

Write the expression for lens maker’s formula.

  $\frac{1}{f_a}=\left(\frac{n}{n_a}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ …… (6)

Here, $f_a$ is the focal length of lens in air.

Rearrange above equation for the value of $\frac{1}{R_1}-\frac{1}{R_2}$ .

  $\begin{array}{c}\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{\left(\frac{n}{n_a}-1\right)f_a}\\ =\frac{n_a}{\left(n-n_a\right)f_a}\end{array}$

Substitute $\frac{n_a}{\left(n-n_a\right)f_a}$ for $\frac{1}{R_1}-\frac{1}{R_2}$ in equation (5).

  $\frac{n_w}{f_w}=\left(n-n_w\right)\left(\frac{n_a}{\left(n-n_a\right)f_a}\right)$

Rearrange above equation for the value of $f_w$ .

  $f_w=f_a\left(\raisebox{1ex}{$n_w$}\!\left/ \!\raisebox{-1ex}{$n_a$}\right.\right)\left(\raisebox{1ex}{$\left(n-n_a\right)$}\!\left/ \!\raisebox{-1ex}{$\left(n-n_w\right)$}\right.\right)$

Conclusion:

Thus, the focal length of a thin lens in water is given by $f_w=f_a\left(\raisebox{1ex}{$n_w$}\!\left/ \!\raisebox{-1ex}{$n_a$}\right.\right)\left(\raisebox{1ex}{$\left(n-n_a\right)$}\!\left/ \!\raisebox{-1ex}{$\left(n-n_w\right)$}\right.\right)$ .

(b)

To determine

The focal length of a double concave lens in air and water.

Explanation of Solution

Given:

Double concave lens has an index of refraction of $1.50$ and radii of magnitude are $30\text{cm}$ and $35\text{cm}$ .

Formula used:

Write the expression for lens maker’s formula.

  $\frac{1}{f_a}=\left(\frac{n}{n_a}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Here, $f_a$ is the focal length of lens in air, $n_a$ is the refractive index of air, $n$ is the refractive index of glass of lens, $R_1$ and $R_2$ are the radii of curvature of lens.

Calculation:

Substitute $1.50$ for $n$ , $1$ for $n_a$ , $-30\text{cm}$ for $R_1$ and $35\text{cm}$ for $R_2$ in equation (2).

  $\begin{array}{c}\frac{1}{f_a}=\left(\frac{1.5}{1}-1\right)\left(\frac{1}{-30\text{cm}}-\frac{1}{35\text{cm}}\right)\\ =-\frac{1}{32\text{cm}}\end{array}$

Rearrange above equationfor $f_a$ .

  $f_a=-32\text{cm}$

Write the expression for the focal length of a thin lens in water.

  $f_w=f_a\left(\raisebox{1ex}{$n_w$}\!\left/ \!\raisebox{-1ex}{$n_a$}\right.\right)\left(\raisebox{1ex}{$\left(n-n_a\right)$}\!\left/ \!\raisebox{-1ex}{$\left(n-n_w\right)$}\right.\right)$ .

Here, $f_w$ is the focal length of lens in water, $f_a$ is the focal length of lens in air, $n_a$ is the refractive index of air, $n$ is the refractive index of glass of lens and $n_w$ is the refractive index of water.

Substitute $1.33$ for $n_w$ , $1.50$ for $n$ , $1$ for $n_a$ and $-32\text{cm}$ for $f_a$ in above equation.

  $\begin{array}{c}{f}_w=\frac{\left(1.33\right)\left(1.5-1\right)}{\left(1.5-1.33\right)}\left(-32\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\\ =-1.3\text{m}\end{array}$

Conclusion:

Thus, the focal length of lens in air is $-32\text{cm}$ and the focal length of lens in water is $-1.3\text{m}$ .