An object of height 6 cm is placed in front of a divergent lens of focal length 150 cm, the object is 300 cm from the lens. Determine:
a) The position and height of the image;
b) The magnification of the image.


Problem data:

• Object height:    o = 6 cm;
• Lens focal length:    f = −150 cm
• Distance from object to lens:    p = 300 cm.

Image construction:

Using a sign convention, the horizontal direction from which the light ray come is positive (the left, where the object is) and upwards in the vertical direction (Figure 1).


Using the rule, all rays of light that are parallel to the principal axis pass through the focal point F', we have a ray that comes out of the object and hits the lens, connecting this point with F', and we have the direction with which the ray will leave the other side of the lens (Figure 2).


Using the rule, all rays of light incident at the center of the lens continue in a straight line, from the crossing of this ray with the extension of the ray previously obtained we have the position of the image i (Figure 3).


Problem diagram:


Solution

a) As the lens is divergent   f < 0, the distance from the image (abscissa) is calculated by the Thin Lens Equation multiplying and dividing by 2, the first term on the right-hand side of the equation The size of the image is given by the Magnification Equation substituting the values ​​in the last equation What agrees with the problem diagram, the image is virtual  p' < 0 and smaller than the object.

b) The magnification is obtained by applying the first equality of the above equation The image has \( \dfrac{1}{3} \) of the size of the object.