Question 23.8: The Case of a Diverging Lens GOAL Calculate geometric quanti...
The Case of a Diverging Lens
GOAL Calculate geometric quantities associated with a diverging lens.
PROBLEM Repeat the problem of Example 23.7 for a diverging lens of focal length 10.0 cm.
STRATEGY Once again, substitution into the thin-lens equation and the associated magnification equation, together with the conventions in Table 23.3, solve the various parts. The only difference is the negative focal length.
| Table 23.3 Sign Conventions for Thin Lenses | |||||
| Quantity | Symbol | In Front | In Back | Convergent | Divergent |
| Object location | + | – | |||
| Image location | q | – | + | ||
| Lens radii |
R_{1},R_{2} |
– | + | ||
| Focal length | + | – | |||
(a) Locate the image and its magnification if the object is at 30.0 cm.
The ray diagram is given in Figure 23.27a. Apply the thin-lens equation with p = 30.0 cm to locate the image:
{\frac{1}{p}}+{\frac{1}{q}}={\frac{1}{f}}
{\frac{1}{30.0\,\mathrm{cm}}}+{\frac{1}{q}}=-\,{\frac{1}{10.0\,\mathrm{cm}}}
Solve for q, which is negative and hence virtual:
q = -7.50 cm
Substitute into Equation 23.10 to get the magnification.
Because M is positive and has absolute value less than 1, the image is upright and smaller than the object:
M={}-\frac{q}{p}=\,-\left(\frac{-7.50\,\mathrm{cm}}{30.0\,\mathrm{cm}}\right)=\;{+}~0.250
M=\frac{h^{\prime}}{h}=-\frac{q}{p} [23.10]
(b) Locate the image and find its magnification if the object is 10.0 cm from the lens.
Apply the thin-lens equation, taking p = 10.0 cm:
{\frac{1}{10.0\,\mathrm{cm}}}+{\frac{1}{q}}=-\,{\frac{1}{10.0\,\mathrm{cm}}}
Solve for q (once again, the result is negative, so the image is virtual):
q = -5.00 cm
Calculate the magnification. Because M is positive and has absolute value less than 1, the image is upright and smaller than the object:
M=-\,\frac{q}{p}=\,-\left(\frac{-5.00\,\mathrm{cm}}{10.0\,\mathrm{cm}}\right)=\;{+\;0.500}
(c) Locate the image and find its magnification when the object is at 5.00 cm.
The ray diagram is given in Figure 23.27b. Substitute p = 5.00 cm into the thin-lens equation to locate the image:
{\frac{1}{5.00\ {\mathrm{cm}}}}+{\frac{1}{q}}=-\,{\frac{1}{10.0\ {\mathrm{cm}}}}
Solve for q. The answer is negative, so once again the image is virtual:
q = -3.33 cm
Calculate the magnification. Because M is positive and less than 1, the image is upright and smaller than the object:
{M}=-\bigg(\frac{-3.33\,\mathrm{cm}}{5.00\,\mathrm{cm}}\bigg)= + 0.666
REMARKS Notice that in every case the image is virtual, hence on the same side of the lens as the object. Further, the image is smaller than the object. For a diverging lens and a real object, this is always the case, as can be proven mathematically.
