Question

Suppose your \(50.0{\rm{ }}mm\) focal length camera lens is \(51.0{\rm{ }}mm\) away from the film in the camera.

(a) How far away is an object that is in focus?

(b) What is the height of the object if its image is \(2.00{\rm{ }}cm\) high?

Short answer

(a) The distance of the object in focus from the camera is\({d_o} = 2550\;mm\).

(b) If the image of an object is \(2.00\;cm\) high then its height is \({h_o} = - 1000\;mm\).

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 where two parallel light beams meet or converge. Depending on the nature of the lens and mirror (concave or convex), the focal length varies with the sign (positive or negative).

Information Provided

• Focal length of the lens:\(50.0\;mm = 50 \times {10^{ - 3}}\;m\).
• Distance between lens and film:\(51.0\;mm = 51 \times {10^{ - 3}}\;m\).
• Size of the image: \(2.00\;cm = 0.02\;m\).

Distance of the object

(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 \({d_i}\) in this equation to solve for \({d_o}\).

Rearranging the expression and substituting the values –

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

\(\begin{array}{c}{d_o} = {\left( {\frac{1}{{50.0 \times {{10}^{ - 3}}\;m}} - \frac{1}{{51.0 \times {{10}^{ - 3}}\;m}}} \right)^{ - 1}}\\ = 2.55\;m\\ = 2.55 \times 1000\;mm\\ = 2550\;mm\end{array}\)

Therefore, the value for distance is obtained as \({d_o} = 2550\;mm\).

Height of the object

Step 4: Height of the object

(b)

The magnification of the lens is given by the relation:

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

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.

Rearranging and solving for\({h_o}\), it is obtained –

\({h_o} = - \frac{{{d_o}}}{{{d_i}}} \times {h_i}\)

\(\begin{array}{c} = \frac{{ - 2.55\;m}}{{0.051\;m}} \times 0.02\;m\\ = - 1.00\;m\\ = - 1.00 \times 1000\;mm\\ = - 1000\;mm\end{array}\)

Therefore, the value for height is obtained as \({h_o} = - 1000\;mm\).