Question 23.5: Gaze into the Crystal Ball Goal Calculate the properties of ...
Gaze into the Crystal Ball
Goal Calculate the properties of an image created by a spherical lens.
Problem A coin 2.00 \mathrm{~cm} in diameter is embedded in a solid glass ball of radius 30.0 \mathrm{~cm} (Fig. 23.17). The index of refraction of the ball is 1.50 , and the coin is 20.0 \mathrm{~cm} from the surface. (a) Find the position of the image of the coin, and (b) the height of the coin’s image.
Strategy Because the rays are moving from a medium of high index of refraction (the glass ball) to a medium of lower index of refraction (air), those originating at the coin are refracted away from the normal at the surface and diverge outward. The image is formed in the glass and is virtual. Substitute into Equations 23.7 and 23.8
{\frac{n_{1}}{p}}+{\frac{n_{2}}{q}}={\frac{n_{2}-\,n_{1}}{R}} (23.7)
M={\frac{h^{\prime}}{h}}=-{\frac{n_{1}q}{n_{2}\rho}} (23.8)
for the image position and magnification, respectively.
Apply Equation 23.7, and take n_{1}=1.50, n_{2}=1.00, p=20.0 \mathrm{~cm}, and R=-30.0 \mathrm{~cm} :
\begin{aligned} \frac{n_{1}}{p}+\frac{n_{2}}{q} & =\frac{n_{2}-n_{1}}{R} \\ \frac{1.50}{20.0 \mathrm{~cm}}+\frac{1.00}{q} & =\frac{1.00-1.50}{-30.0 \mathrm{~cm}} \end{aligned}
Solve for q :
q=-17.1 \mathrm{~cm}
To find the image height, we use Equation 23.8 for the magnification:
\begin{aligned} & M=-\frac{n_{1} q}{n_{2} p}=-\frac{1.50(-17.1 \mathrm{~cm})}{1.00(20.0 \mathrm{~cm})}=\frac{h^{\prime}}{h} \\ & h^{\prime}=1.28 h=(1.28)(2.00 \mathrm{~cm})=2.56 \mathrm{~cm} \end{aligned}
Remarks The negative sign on q indicates that the image is in the same medium as the object (the side of incident light), in agreement with our ray diagram, and therefore must be virtual. The positive value for M means the image is upright.