Example 13.3: Converging lenses

Question: An object of height $h=7$cm is placed a distance $p=25$cm in front of a thin converging lens of focal length $f=35$cm. What is the height, location, and nature of the image? Suppose that the object is moved to a new location a distance $p=90$cm in front of the lens. What now is the height, location, and nature of the image?
 
Answer: According to Eq. (367), the image distance $q$ is given by

$$q = \frac{1}{1/f - 1/p} = \frac{1}{(1/35-1/25)} = - 87.5\,{\rm cm}.$$

Thus, the image is virtual (since $q<0$), and is located $87.5$cm in front of the lens. According to Eq. (10.24), the magnification $M$ of the image is given by

$$M = -\frac{q}{p} = -\frac{(-87.5)}{(25)} = 3.5.$$

Thus, the image is upright (since $M>0$), and magnified by a factor of $3.5$. It follows that the height $h'$ of the image is given by

$$h' = M\,h =(3.5)\,(7) =24.5\,{\rm cm}.$$

If the object is moved such that $p=\,90$cm then the new image distance is given by

$$q = \frac{1}{1/f - 1/p} = \frac{1}{(1/35-1/90)} = 57.27\,{\rm cm}.$$

Thus, the new image is real (since $q>0$), and is located 57.27cm behind the lens. The new magnification is given by

$$M = -\frac{q}{p} = -\frac{(57.27)}{(90)} = -0.636.$$

Thus, the image is inverted (since $M<0$), and diminished by a factor of $0.636$. It follows that the new height of the image is

$$h' = M\,h = -(9.636)\,(7) = -4.45\,{\rm cm}.$$

Note that the height is negative because the image is inverted.