Question 23.9: Two Lenses in a Row GOAL Calculate geometric quantities for ...

(a) Locate the image and determine the magnification of lens 1.
See the ray diagram, Figure 23.28b. Apply the thin-lens equation to lens 1:

{\frac{1}{30.0\ {\mathrm{cm}}}}+{\frac{1}{q}}={\frac{1}{10.0\ {\mathrm{cm}}}}

Solve for q, which is positive and hence to the right of the first lens:

q = + 15.0 cm

Compute the magnification of lens 1:

M_{1}=-\,{\frac{q}{p}}=-\,{\frac{15.0\,\mathrm{cm}}{30.0\,\mathrm{cm}}}=\,-~0.500

(b) Locate the final image and find the total magnification.
The image formed by lens 1 becomes the object for lens 2.
Compute the object distance for lens 2:

p=20.0\;\mathrm{cm}-15.0\;\mathrm{cm}=5.00~\mathrm{cm}

Once again apply the thin-lens equation to lens 2 to locate the final image:

{\frac{1}{5.00  \mathrm{m}}}+{\frac{1}{q}}={\frac{1}{20.0\,\mathrm{cm}}}

q = -6.67 cm

Calculate the magnification of lens 2:

M_{2}=-\,{\frac{q}{p}}=\,-\,{\frac{\,(-6.67\,\mathrm{cm})\,}{5.00\,\mathrm{cm}}}=\,+1.33

Multiply the two magnifications to get the overall magnification of the system:

M=M_{1}M_{2}=(-0.500)(1.33)={\mathrm{-0.665}}

REMARKS The negative sign for M indicates that the final image is inverted and smaller than the object because the absolute value of M is less than 1. Because q is negative, the final image is virtual.