Question

An old-fashioned slide projector needs to create a 98-cm-high image of a 2.0-cm-tall slide. The screen is 300 cm from the slide.

a. What focal length does the lens need? Assume that it is a thin lens.

b. How far should you place the lens from the slide?

Short answer

a) Focal length of the lens is 5.9 cm

b) The lens is located at a distance of 6.0 cm from the slide and 294 cm from the screen.

Step-by-step solution

Step 1. Given information is :

Height of the image, h' = 98 cm

Height of the slide, h = 2 cm

Distance between the slide and the screen = 300 cm

We need to find out :

a) Focal length of the lens

b)The distance s' between slide and the lens

Step 2. Using Magnification concept,

$$\begin{gathered} \left|M\right|=\frac{h\text{'}}{h}h\text{'}andhareimageandobjectheightrespectively. \\ \left|M\right|=\frac{98cm}{2.0cm}=49 \end{gathered}$$ M is the Magnification.

Step 3. Magnification in terms of object and image distance 

$$\begin{gathered} M=-\frac{s\text{'}}{s} \\ wheres\text{'}andsareimageandobjectdis\tan cerespectively. \\ Itisgiventhats+s\text{'}=300cm \\ s\text{'}=300cm-s \\ M=-\frac{300cm-s}{s} \\ -49=-\frac{300cm-s}{s} \\ 49=\frac{300cm-s}{s} \\ 49s=300cm-s \\ 50s=300cm \\ s=\frac{300cm}{50}=6.0cm \\ Ands\text{'}=300cm-s=300cm-6.0cm=294cm \end{gathered}$$

Step 4. Using thin lens formula

$$\begin{gathered} \frac{1}{s}+\frac{1}{s\text{'}}=\frac{1}{f} \\ \frac{1}{f}=\frac{s\text{'}+s}{ss\text{'}} \\ f=\frac{ss\text{'}}{s\text{'}+s} \\ f=\frac{(6.0cm)(294cm)}{6.0cm+294cm} \\ f=5.9cm \end{gathered}$$