Chapter 8.5, Problem 37PPSE

Solution

To calculate:The equation for the unknown variable and units used for the focal length of the lens if for the image of the overhead projector to be in focus, the distance $d_i$ from projector lens to the image, the projector lens’s focal length $f$ and distance $d_0$ from transparency to the projector lens must satisfy the thin lens equation $\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_0}$ .

The equation for the unknown variable is $f=\frac{d_i{d}_0}{d_0+d_i}$ and units used for the focal length of the lens is $3.84$ inches.

Given information:

The image of the overhead projector to be in focus, the distance $d_i$ from projector lens to the image, the projector lens’s focal length $f$ and distance $d_0$ from transparency to the projector lens must satisfy the thin lens equation $\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_0}$ .

Formula Used:

To simplify the sum or difference of rational expressions, look for the LCD of the expression. Then create an equivalent expression using the LCD and combine the like terms.

Calculation:

Take $d_0$ to be $4$ in and $d_i$ to be $8$ ft that is $8\cdot 12=96$ inches.

Now,

  $\frac{1}{f}=\frac{1}{96}+\frac{1}{4}$

Multiply each side with LCD $96f$ .

  $\begin{array}{c}96f\cdot \frac{1}{f}=\left(\frac{1}{96}+\frac{1}{4}\right)\cdot 96f\\ 96=f+24f\\ 96=25f\\ f=\frac{96}{25}\end{array}$

  $f=3.84$ inches

Now, to find an equation for the unknown variable $f$ .

Multiply each side of equation $\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_0}$ with LCD $f{d}_i{d}_0$ .

  $\begin{array}{c}f{d}_i{d}_0\frac{1}{f}=\left(\frac{1}{d_i}+\frac{1}{d_0}\right)f{d}_i{d}_0\\ d_i{d}_0=f{d}_0+f{d}_i\\ d_i{d}_0=f\left(d_0+d_i\right)\\ f=\frac{d_i{d}_0}{d_0+d_i}\end{array}$

Thus, the equation for the unknown variable is $f=\frac{d_i{d}_0}{d_0+d_i}$ and units used for the focal length of the lens is $3.84$ inches.