Chapter 23, Problem 54AP

Two rays travelling parallel to the principal axis strike a large plano-convex lens having a refractive index of 1.60 (Fig. P23.54). If the convex face is spherical, a ray near the edge does not pass through the local point (spherical aberration occurs). Assume this face has a radius of curvature of R = 20.0 cm and the two rays are at distances h₁ = 0.500 cm and h₂ = 12.0 cm from the principal axis. Find the difference Δx in the position where each crosses the principal axis.

Figure P23.54

To determine

The distance between the points where the two rays cross the principle axis.

Answer to Problem 54AP

The distance between the points where the two rays cross the principle axis is $21.3\text{cm}$.

Explanation of Solution

Given info:

The radius of curvature is $20.0\text{cm}$.

The refractive index of the lens is $1.60$.

The refractive index of air is $1.00$.

The distance the first light ray from the principal axis is $0.500\text{cm}$.

The distance the second light ray from the principal axis is $12.0\text{cm}$.

The following diagram shows the light ray travelling parallel to the principle axis of a plano- convex lens.

The ray strikes the plane surface at normal incidence and passes into the glass undeviated.

Formula to calculate the angle of incidence on the spherical surface is,

${\theta }_1={\sin }^{-1}\left(\frac{h}{R}\right)$

• ${\theta }_1$ is the angle of incidence
• $h$ is the height of lens
• $R$ is the radius of curvature

Formula to calculate the angle of refraction at the spherical surface is,

$\begin{array}{c}{\theta }_2={\sin }^{-1}\left(\frac{n\sin {\theta }_1}{n_{air}}\right)\\ ={\sin }^{-1}\left(\frac{n\sin \left({\sin }^{-1}\left(\frac{h}{R}\right)\right)}{n_{air}}\right)\\ ={\sin }^{-1}\left(\frac{nh}{n_{air}R}\right)\end{array}$

• ${\theta }_2$ is the angle of refraction
• $n$ is the refractive index of lens
• $n_{air}$ is the refractive index of air

Formula to calculate the distance from the center of the spherical surface to the point where the refracted ray crosses the principle axis of the lens is,

$\begin{array}{c}x=\sqrt{R^2-h^2}+\frac{h}{\tan \left({\theta }_2-{\theta }_1\right)}\\ =\sqrt{R^2-h^2}+\frac{h}{\tan \left(\left({\sin }^{-1}\left(\frac{nh}{n_{air}R}\right)\right)-\left({\sin }^{-1}\left(\frac{h}{R}\right)\right)\right)}\end{array}$

• $x$ is the distance from the center of the spherical surface to the point where the refracted ray crosses the principle axis of the lens

Formula to calculate the distance from the center of the spherical surface to the point where the refracted ray for the first incoming ray crosses the principle axis of the lens is,

$x_1=\sqrt{R^2-h_1^2}+\frac{h_1}{\tan \left(\left({\sin }^{-1}\left(\frac{n{h}_1}{n_{air}R}\right)\right)-\left({\sin }^{-1}\left(\frac{h_1}{R}\right)\right)\right)}$

• $x_1$ is the distance from the center of the spherical surface to the point where the refracted ray for the first incoming ray crosses the principle axis of the lens
• $h_1$ is the distance the first light ray from the principal axis

Formula to calculate the distance from the center of the spherical surface to the point where the refracted ray for the second incoming ray crosses the principle axis of the lens is,

$x_2=\sqrt{R^2-h_2^2}+\frac{h_2}{\tan \left(\left({\sin }^{-1}\left(\frac{n{h}_2}{n_{air}R}\right)\right)-\left({\sin }^{-1}\left(\frac{h_2}{R}\right)\right)\right)}$

• $x_2$ is the distance from the center of the spherical surface to the point where the refracted ray for the second incoming ray crosses the principle axis of the lens
• $h_2$ is the distance the second light ray from the principal axis

Formula to calculate the distance between the points where the two rays cross the principle axis is,

$\begin{array}{c}\Delta x=x_1-x_2\\ =\left(\begin{array}{c}\left(\sqrt{R^2-h_1^2}+\frac{h_1}{\tan \left(\left({\sin }^{-1}\left(\frac{n{h}_1}{n_{air}R}\right)\right)-\left({\sin }^{-1}\left(\frac{h_1}{R}\right)\right)\right)}\right)-\\ \left(\sqrt{R^2-h_2^2}+\frac{h_2}{\tan \left(\left({\sin }^{-1}\left(\frac{n{h}_2}{n_{air}R}\right)\right)-\left({\sin }^{-1}\left(\frac{h_2}{R}\right)\right)\right)}\right)\end{array}\right)\end{array}$

• $\Delta x$ is the distance between the points where the two rays cross the principle axis

Substitute $1.00$ for $n_{air}$, $1.60$ for $n$, $0.500\text{cm}$ for $h_1$, $12.0\text{cm}$ for $h_2$ and $20.0\text{cm}$ for $R$ to find $\Delta x$.

$\begin{array}{c}\Delta x=\left(\begin{array}{c}\left(\sqrt{{\left(20.0\text{cm}\right)}^2-{\left(0.500\text{cm}\right)}^2}+\frac{\left(0.500\text{cm}\right)}{\tan \left(\left({\sin }^{-1}\left(\frac{\left(1.60\right)\left(0.500\text{cm}\right)}{\left(1.00\right)\left(20.0\text{cm}\right)}\right)\right)-\left({\sin }^{-1}\left(\frac{0.500\text{cm}}{20.0\text{cm}}\right)\right)\right)}\right)-\\ \left(\sqrt{{\left(20.0\text{cm}\right)}^2-{\left(12.0\text{cm}\right)}^2}+\frac{\left(12.0\text{cm}\right)}{\tan \left(\left({\sin }^{-1}\left(\frac{\left(1.60\right)\left(12.0\text{cm}\right)}{\left(1.00\right)\left(20.0\text{cm}\right)}\right)\right)-\left({\sin }^{-1}\left(\frac{12.0\text{cm}}{20.0\text{cm}}\right)\right)\right)}\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(\sqrt{{\left(20.0\text{cm}\right)}^2-{\left(0.500\text{cm}\right)}^2}+\frac{\left(0.500\text{cm}\right)}{\tan \left(2.29°-1.43°\right)}\right)-\\ \left(\sqrt{{\left(20.0\text{cm}\right)}^2-{\left(12.0\text{cm}\right)}^2}+\frac{\left(12.0\text{cm}\right)}{\tan \left(73.7°-36.9°\right)}\right)\end{array}\right)\\ =53.3\text{cm}-32.0\text{cm}\\ =21.3\text{cm}\end{array}$

Thus, the distance between the points where the two rays cross the principle axis is $21.3\text{cm}$

Conclusion:

The distance between the points where the two rays cross the principle axis is

$21.3\text{cm}$.