Set 56 Problem number 22

Problem

A person 1.47 meters high stands 5 meters from a converging lens with focal length 60 millimeters. 

• Where will the image form? 
• Will the image be upright or inverted? 
• How large will the image be? 
• Will it be real or virtual?  
• Sketch a diagram explaining how the image forms.

Solution

The object distance is dObj = 5 meters, and the focal length is f = 9 m.   If the image distance is dImg, we have

• 1 / dObj + 1/  dImg = 1 / f.

We solve this equation for dImg, obtaining

• dImg = f * dObj / (dObj + f) = 9 m * 5 m / ( 9 m + 5 m) = 0 meters.

The ratio of image size hImg to object size hObj is the same as the ratio of image distance to object distance, or

• hImg / hObj = g m / (b m) = 0,

so

• hImg = hObj * 0 = 5 meters * 0 = 0 meters.

The ray diagram below shows that the principal ray from the top of the person's head, directed parallel to the axial ray, will pass through the lens and be deflected in the direction of the focal point on the other side. 

• The central ray from the top of the head will pass effectively undeflected through a thin lens. 
• Since the persons stands outside the focal point of the lens, this central ray will intersect the first ray. 
• The intersection point will be just past the focal point, at the previously calculated 0 meter distance. 
• At this point both rays will be below the axial ray, so the image will be inverted.

Note that a similar diagram could show the principle rays reflected from any point on the object.  These rays will intersect on the corresponding point of the image.  All the other rays reflected from the same point on the object and passing through the lens will intersect at the same point on the image.  Thus a screen placed at the position of the image can show the image--the image is actually there, since the rays really focus at the position of the image.