Question

An object of height 2.7 cm is placed 29 cm in front of a diverging lens of focal length 16 cm. Behind the diverging lens, and 12 cm from it, there is a converging lens of the same focal length.
A. Find the location of the final image, in centimeters beyond the converging lens.
B. What is the magnification of the final image?

Answer 1

Answer:

A)  q = -8.488 cm ,  B)  m = 0.29

Explanation:

A) For this exercise in geometric optics, we will use the equation of the constructor

          $\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where p and q are the distance to the object and image, respectively and f is the focal length

in our case the distance the object is p = 29 cm the focal length of a diverging lens is negative and indicates that it is f = - 12 cm

         $\frac{1}{q} = \frac{1}{f} - \frac{1}{p}$

     

we calculate

          $\frac{1}{q} = - \frac{1}{12} - \frac{1}{29}$

          $\frac{1}{q}$ = - 0.1178

          q = -8.488 cm

the negative sign indicates that the image is virtual

B) the magnification is given

          $m = \frac{h'}{h} = - \frac{q}{p}$

       

we substitute

          m = $- \frac{-8.488}{29}$

          m = 0.29

the positive sign indicates that the image is right