Solved Problem on Spherical Mirrors

Determine the position and height of the image formed by a spherical mirror with a radius of 60 cm, with an object of height 3 cm situated 20 cm from the vertex of a convex mirror.

Problem data:

Radius of curvature of the mirror: R = 60 cm;
Height of the object: o = 3 cm;
Distance from the object to the vertex of the mirror: p = 20 cm.

Diagram of the problem

Using a sign convention, where the horizontal direction from which the light ray comes (to the left, where the object is) is positive and upwards in the vertical direction (Figure 1).

Figure 1

The distance from the focus to the vertex, f, will be half the radius of curvature, R, since the mirror is convex, its focus is negative (f < 0)

\[ \begin{gather} \bbox[#99CCFF,10px] {f=-{\frac{R}{2}}} \end{gather} \]

\[ \begin{gather} f=-{\frac{60\;\mathrm{cm}}{2}}\\[5pt] f=-30\;\mathrm{cm} \tag{I} \end{gather} \]

Construction of image of the convex mirror

Drawing one first ray of light using the rule of the spherical mirrors that states, every ray of light incident parallel to the principal axis is reflected passing through the principal focus of the mirror (Figure 2).

Figure 2

Drawing a second ray using the rule that states, every rays of light incident on the vertex of the mirror reflects simmetrically to the principal axis (Figure 3). Since there is no crossing of the rays reflected in front of the mirror, we see that they intersect behind the mirror where the image is formed.

Figure 3

Figure 4

Solution

To find the distance of the image to the mirror, q, we use the Mirror Equation

\[ \begin{gather} \bbox[#99CCFF,10px] {\frac{1}{f}=\frac{1}{p}+\frac{1}{q}} \end{gather} \]

using the distance from the object to the mirror, p, given in the problem and the focal distance, f, obtained in expression (I)

\[ \begin{gather} \frac{1}{-30\;\mathrm{cm}}=\frac{1}{20\;\mathrm{cm}}+\frac{1}{q}\\[5pt] -{\frac{1}{30\;\mathrm{cm}}}=\frac{1}{20\;\mathrm{cm}}+\frac{1}{q}\\[5pt] \frac{1}{q}=-{\frac{1}{30\;\mathrm{cm}}}-\frac{1}{20\;\mathrm{cm}} \end{gather} \]

the Least Common Multiple (LCM) between 30 and 20 is 60

\[ \begin{gather} \frac{1}{q}=\frac{-2\;\mathrm{cm}-3\;\mathrm{cm}}{60\;\mathrm{cm^2}}\\[5pt] \frac{1}{q}=\frac{-5\;\mathrm{\cancel{cm}}}{60\;\mathrm{cm^{\cancel 2}}}\\[5pt] q=-{\frac{60\;\mathrm{cm}}{5}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {q=-12\;\mathrm{cm}} \end{gather} \]

The size of the image, i, is given by Magnification Equation

\[ \begin{gather} \bbox[#99CCFF,10px] {\frac{i}{o}=-{\frac{q}{p}}} \end{gather} \]

using the distance from the object to the mirror and the height of the object given in the problem, and the distance from the image to the mirror obtained above

\[ \begin{gather} \frac{i}{3\;\mathrm{cm}}=-{\frac{(-12\;\mathrm{cm})}{20\;\mathrm{cm}}}\\[5pt] \frac{i}{3\;\mathrm{cm}}=\frac{12\;\mathrm{cm}}{20\;\mathrm{cm}}\\[5pt] i=\frac{(3\;\mathrm{cm})(12\;\mathrm{cm})}{20\;\mathrm{cm}}\\[5pt] i=\frac{36\;\mathrm{cm^{\cancel 2}}}{20\;\mathrm{\cancel{cm}}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {i=1.8\;\mathrm{cm}} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .