Question

A certain slide projector has a \(100{\rm{ }}mm\) focal length lens.

(a) How far away is the screen, if a slide is placed \(103{\rm{ }}mm\) from the lens and produces a sharp image?

(b) If the slide is \(24.0\) by \(36.0{\rm{ }}mm\), what are the dimensions of the image? Explicitly show how you follow the steps in the Problem-Solving Strategy for lenses.

Short answer

(a) If the slide is placed\(103\;mm\)away from the lens and produces a sharp image, the screen is at a distance\({d_i} = 3430\;mm\).

(b) If the slide is \(24\;mm\) by \(36\;mm\), the dimensions of the image are \(799\;mm\) by \(1200\;mm\).

Step-by-step solution

Concept Introduction

The focal length is the distance between a convex lens or a concave mirror and the focal point of a lens or mirror. It is the point at which two parallel light beams meet or converge. Depending on the lens and mirror (concave or convex), the focal length varies with the sign (positive or negative).

Information Provided

• The focal length of the projector slide:\(100\;mm = 100 \times {10^{ - 3}}\;m\).
• Distance of slide from the lens:\(103\;mm = 103 \times {10^{ - 3}}\;m\).
• Dimension of the slide: \(24\;mm,{\rm{ }}36\;mm = 0.024\;m,{\rm{ }}0.036\;m\)

Distance of the screen

(a)

Use the thin lens equation for the focal length:

\(\frac{1}{f} = \frac{1}{{{d_i}}} + \frac{1}{{{d_o}}}\)

where \(f\) denotes the lens's focal length, \({d_o}\) denotes the distance between the object and the lens, and \({d_i}\) denotes the distance from the lens to the projected image that is in focus.

Substitute with the given values of \(f\) and \({d_o}\) in this equation to solve for \({d_i}\).

Rearranging the expression and solving –

\(\frac{1}{{{d_i}}} = \frac{1}{f} - \frac{1}{{{d_o}}}\)

\(\begin{array}{c}{d_i} = {\left( {\frac{1}{{100 \times {{10}^{ - 3}}\;\;m}} - \frac{1}{{103 \times {{10}^{ - 3}}\;\;m}}} \right)^{ - 1}}\\ = 3.43\;\;m\\ = 3.43 \times 1000\;mm\\ = 3430\;mm\end{array}\)

Therefore, the value for the distance is obtained as \({d_i} = 3430\;mm\).

Dimensions of the image

(b)

Use the equation of the magnification that relates the image size to the object size: \(m = \frac{{{h_i}}}{{{h_o}}}\)with that which relates the image distance to the object distance:

\(m = - \frac{{{d_i}}}{{{d_o}}}\).

where \(m\) denotes the magnification, \({d_o}\) denotes the distance between the object and the lens, \({d_i}\) denotes the distance from the lens to the projected image that is in focus, denotes \({h_i}\) height of reflection, and reflects \({h_o}\) height of object.

The image and object distances are both given in part (a); use them with the given dimensions (call them \({h_{o1}}\) and \({h_{o2}}\)) of the slide to solve for the image dimensions (call them \({h_{i1}}\) and \({h_{i2}}\)).

\(\begin{array}{c}m = - \frac{{{d_i}}}{{{d_o}}}\\ = \frac{{{h_i}}}{{{h_o}}} = \left| { - \frac{{{d_i}}}{{{d_o}}}} \right|\\{h_i} = \frac{{{d_i}}}{{{d_o}}} \times {h_o}\end{array}\)

The calculation for the first dimension is –

\(\begin{array}{c}{h_{i1}} = \frac{{{d_i}}}{{{d_o}}} \times {h_{o1}}\\ = \frac{{3.43\;\;m}}{{0.103\;m}} \times 0.024\;m\\ = 0.799\;m\;\\ = 0.799 \times 1000\;mm\\ = 799\;mm\end{array}\)

The calculation for the second dimension is –

\(\begin{array}{c}{h_{i2}} = \frac{{{d_i}}}{{{d_o}}} \times {h_{o2}}\\ = \frac{{3.43\;m}}{{0.103\;m}} \times 0.036\;m\\ = 1.20\;m\\ = 1.20 \times 1000\;m\\ = 1200\;mm\end{array}\)

Therefore, the values for dimensions are \({h_{i1}} = 799\;mm\) and \({h_{i2}} = 1200\;mm\).