Question

An object 5 centimeters high is placed 15 cm in front of a converging lens with focal length 5 cm. Using graph paper (turn the paper so that the long side is horizontal) and a straight edge, draw appropriate rays to locate the image. Chose a scale (tell me what it is!) such that the lens is in the center of the page and the object is near the left side of the page a. Is the image real or imaginary? Explain your conclusion b. What is the height of the image? c. Where is the image located? d. What is the ratio (image height)/(object height)? e. Compare your results with those predicted by the lens equation 1/f = 1/p + 1/q

Answer 1

The image formation formed in the graph paper is shown below.

The scale taken on plotting the ray diagram is 1 unit on x-axis is 2cm and 1 unit on y-axis is 2cm. From the graph,

Object distance (p) = 7.5 units = 7.5 x 2 = 15cm

Focal length of lens (f) = 2.5 units = 2.5 x 2 = 5cm

Object distance (q) = 3.75 units = 3.75 x 2 = 7.5cm

Magnification (m) is given by

(a) The image formed is real because the magnification is negative.

(b) The height of the image is 1.25units = 1.25 x 2 = 2.5cm.

(c) The image is located at 3.75units = 3.75 x 2 = 7.5cm to the right of the converging lens.

(d) Height of the object (ho) = 2.5 units and height of the image (ho) is 1.25 units

(e) The lens equation is given by,

Given in the question p = 15cm and f = 5cm

So the image is located 7.5cm to the right of the converging lens.

Magnification is given by,

Since the magnification is negative, the image is real.

So the results from the lens equation and the ray diagram are same.