Chapter 25, Problem 47P

To determine

The locations of the images of red and yellow objects, when light from the objects is allowed to pass through a Plano convex lens.

Answer to Problem 47P

Solution:

The image of the red object is formed at 49.8 cm from the lens and the image of the yellow object is formed at a distance 47.7 cm from the lens.

Explanation of Solution

The focal length of a lens is dependent on the wavelength of incident light. The lens maker’s formula is written as,

$\frac{1}{f}=\left(n-1\right)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)$

Here, the focal length of the lens is f, the refractive index of the incident light is n and the radii of curvatures of the two surfaces of the lens are $R_1\text{ and }{R}_2$. Since the lens has different refractive indices for light of different wavelengths, the focal length of the lens also differs for different colours.

The focal length of the lens is determined for the red object and the yellow object. Then, using the thin lens formula, the image distance for each of the objects is calculated.

Given:

The radius of curvature of the plane surface $R_1=\infty$

The radius of curvature of the convex surface $R_2=14.5\text{ cm}$

Object distance $d_o=66.0\text{ cm}$

Refractive index of the lens for red colour $n_r=1.5106$

Refractive index of the lens for yellow colour $n_y=1.5226$

Formula used:

Lens maker’s formula

$\frac{1}{f}=\left(n-1\right)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)$

Thin lens formula

$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$

Calculation:

Calculate the focal length of the lens for the red object using the given quantities of radii of curvatures of the lens surfaces and the refractive index of the lens for red, in the lens maker’s formula.

$\begin{array}{c}\frac{1}{f_r}=\left(n_r-1\right)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\\ =\left(1.5106-1\right)\left(\frac{1}{\infty }+\frac{1}{14.5\text{ cm}}\right)\\ =\frac{0.5106}{14.5\text{ cm}}\end{array}$

Simplifying the expression, the focal length of the lens for red is found to be $f_r=28.4\text{ cm}$.

Use this value of focal length and the given value of the object distance in the thin lens formula and calculate the location of the image of the red object.

$\begin{array}{c}\frac{1}{f_r}=\frac{1}{d_o}+\frac{1}{d_{ir}}\\ \frac{1}{d_{ir}}=\frac{1}{f_r}-\frac{1}{d_o}\\ =\frac{1}{28.4\text{ cm}}-\frac{1}{66.0\text{ cm}}\end{array}$

Simplifying the expression, the image distance for the red object is found to be $d_{ir}=49.8\text{ cm}$.

Repeat the same procedure for the yellow object using the value of the refractive index of the lens for yellow light and calculate the image distance for the yellow object.

For the yellow object, the focal length is calculated as follows:

$\begin{array}{c}\frac{1}{f_y}=\left(n_y-1\right)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\\ =\left(1.5226-1\right)\left(\frac{1}{\infty }+\frac{1}{14.5\text{ cm}}\right)\\ =\frac{0.5226}{14.5\text{ cm}}\\ f_y=27.7\text{ cm}\end{array}$

The image of the yellow object is calculated as follows:

$\begin{array}{c}\frac{1}{f_y}=\frac{1}{d_o}+\frac{1}{d_{iy}}\\ \frac{1}{d_{iy}}=\frac{1}{f_y}-\frac{1}{d_o}\\ =\frac{1}{27.7\text{ cm}}-\frac{1}{66.0\text{ cm}}\\ d_{iy}=47.7\text{ cm}\end{array}$