Equations about Lenses: Solution

Solution

1)  Find the distance of the image

Given:

            F = 30cm

            D_o = 40cm

Equation:  1/f = 1/d_i + 1/d_o

                  1/30 = 1/d_i + 1/40

                  1/d_i = 1/30 – 1/40

                  1/d_i = 1/120

                  d_i = 120cm

2)      Find the magnification

Given:

            D_o = 40cm

            D_i = 120cm

Equation:  Magnification = d_i/d_o

                                         = 120/40

                                         = 3 times

3)      Find the size of the image

Given:

            S_o = 2cm

            D_o = 40cm

            D_i = 120cm

Equation:  s_i/s_o=-d_i/d_o

                 S_id_o = -d_is_o

                 S_i = -(d_is_o)/d_o

                 S_i = -[(120)(2)]/40

                 S_i = -6cm  (The negative sign shows that the image is inverted)

*  An alternative method of finding the size of the image would be to see that the magnification of the object is 3 times.  Knowing that the size of the object was 2cm, the image must be 3 times of this or 6cm tall.

4)      Check to see that your answer makes sense.

The problem gives the focal length as 30cm.  This means that 2F= 60cm.  The object, at 40cm, is between F and 2F.  This is a Case 4 situation.  This means that the image should be magnified, real, inverted, beyond 2F, and on the other side of the lens.  The image is magnified (by 3 times, and it is 6cm tall).  The distance of the image, at 120cm, is definitely beyond 2F (which is 60cm).  We also know that the image is inverted (and therefore real) because we got a negative size for the image.  So our solution checks out correctly.