Question 23.6: The One That Got Away Goal Calculate the properties of an im...
The One That Got Away
Goal Calculate the properties of an image created by a flat refractive surface.
Problem A small fish is swimming at a depth d below the surface of a pond (Fig. 23.18). (a) What is the apparent depth of the fish as viewed from directly overhead? (b) If the fish is 12 \mathrm{~cm} long, how long is its image?
Strategy In this example the refracting surface is flat, so R is infinite. Hence, we can use Equation 23.9
q=-{\frac{n_{2}}{n_{1}}}{{p}} (23.9)
to determine the location of the image, which is the apparent location of the fish.
(a) Find the apparent depth of the fish.
Substitute n_{1}=1.33 for water and p=d into Equation 23.9:
q=-\frac{n_{2}}{n_{1}} p=-\frac{1}{1.33} d=-0.752 d
(b) What is the size of the fish’s image?
Use Equation 23.9 to eliminate q from the Equation 23.8,
M={\frac{h^{\prime}}{h}}=-{\frac{n_{1}q}{n_{2}\rho}} (23.8)
the magnification equation:
\begin{aligned} & M=\frac{h^{\prime}}{h}=-\frac{n_{1} q}{n_{2} p}=-\frac{n_{1}\left(-\frac{n_{2}}{n_{1}} p\right)}{n_{2} p}=1 \\ & h^{\prime}=h=12 \mathrm{~cm} \end{aligned}
Remarks Again, because q is negative, the image is virtual, as indicated in Figure 23.18. The apparent depth is three-fourths the actual depth. For instance, if d=4.0 \mathrm{~m}, then q=-3.0 \mathrm{~m}.