Chapter 1.5, Problem 114E

The Lens Equation If F is the focal length of a convex lens and an object is placed at a distance x from the lens, then its image will be at a distance y from the lens, where F, x, and y are related by the lens equation

$\frac{1}{F}=\frac{1}{x}+\frac{1}{y}$

Suppose that a lens has a focal length of 4.8 cm and that the image of an object is 4 cm closer to the lens than the object itself. How far from the lens is the object?

To determine

The distance of the object from the lens.

Answer to Problem 114E

The distance of the object from the lens is $\underset{\_}{12\text{cm}}$.

Explanation of Solution

Given:

The equation that relates F, x and y is $\frac{1}{F}=\frac{1}{x}+\frac{1}{y}$, where $F$ is the focal length of the convex lens, $x$ is the distance of the object from the lens and $y$ is the  distance of the image from the lens.

Formula used:

Quadratic formula:

The solution of a quadratic equation of the form $a{x}^2+bx+c=0,a\ne 0$ can be obtained by using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Calculation:

The focal length of convex lens is $4.8\text{cm}$.

Since the image of an object is 4 cm closer to the lens than the object itself, it can be written that $y=x-4$.

Substitute $4.8$ for $F$ and $x-4$ for $y$ in the equation $\frac{1}{F}=\frac{1}{x}+\frac{1}{y}$.

$\begin{array}{l}\frac{1}{4.8}=\frac{1}{x}+\frac{1}{x-4}\\ \frac{1}{4.8}=\frac{x-4+x}{x\left(x-4\right)}\\ \frac{1}{4.8}=\frac{2x-4}{x\left(x-4\right)}\\ x\left(x-4\right)=4.8\left(2x-4\right)\end{array}$

Simplify the above equation as follows.

$\begin{array}{l}{x}^2-4x=9.6x-19.2\\ x^2-4x-9.6x+19.2=0\\ x^2-13.6x+19.2=0\end{array}$

Compare this equation with the general form $a{x}^2+bx+c=0$.

Here $a=1,b=-13.6\text{and }c=19$.

Use the quadratic formula to obtain the roots of the given equation.

$\begin{array}{c}x=\frac{-\left(-13.6\right)\pm \sqrt{{\left(-13.6\right)}^2-4\times 1\times 19}}{2\times 1}\\ =\frac{13.6\pm \sqrt{184.96-76}}{2}\\ =\frac{13.6\pm \sqrt{108.96}}{2}\\ =\frac{13.6\pm 10.4}{2}\end{array}$

Simplify further as follows.

$\begin{array}{l}x=\frac{13.6+10.4}{2}\text{or }x=\frac{13.6-10.4}{2}\\ x=\frac{24}{2}\text{or }x=\frac{3.2}{2}\\ x=12\text{or }x=1.6\end{array}$

Substitute $x=12$ in $y=x-4$.

$\begin{array}{c}y=12-4\\ =8\end{array}$

Substitute $x=1.6$ in $y=x-4$.

$\begin{array}{c}y=1.6-4\\ =-2.4\end{array}$

Note that, the distance cannot be negative.

Thus, the distance of the object from the lens is $\underset{\_}{12\text{cm}}$.