Images from Spherical Mirrors — Question 5

Question 5

Image motion can exceed object motion near a concave focus

Use the paraxial mirror relations $1/p+1/q=1/f$ and $m=-q/p$, with p positive for a real object, q positive for a real image, and f positive for a concave mirror. A fixed concave mirror has f=20 cm. At one instant, a real object has p=30 cm and approaches the mirror with dp/dt=-1.0 cm/s.

1. Differentiate the mirror relation to find dq/dp and express it through magnification.

2. Find q and the instantaneous rate dq/dt.

Question 5 - Solution

Image motion can exceed object motion near a concave focus

(a) Method and calculation

$$-\frac1{p^2}-\frac1{q^2}\frac{dq}{dp}=0\Rightarrow\boxed{\frac{dq}{dp}=-\frac{q^2}{p^2}=-m^2}.$$$$\boxed{\dot q=-m^2\dot p}.$$

(b) Interpretation and further calculation

$$\boxed{q=20(30)/(30-20)=60\,\mathrm{cm},\quad m=-2}.$$$$\boxed{\dot q=-4(-1.0)=+4.0\,\mathrm{cm/s}}.$$

Physical check

The positive image-distance rate means the real image recedes from the fixed mirror as the object approaches its focus.