Answer to Question #204891 in Physics for JULIANA

Question

•QUEST A: Search for the tent “Hall of Mirrors” near the faculty of engineering and enter. Look for a light emitter, a piece of paper and pen, a rook chess piece and a converging mirror to answer this question: A rook chess piece is located 2.5 times the focal distance from the converging mirror. If f=5 cm and h_o=5.5 cm, determine the image characteristics (location, object, size, type). If you apply the same distance and location of the object in front of the diverging side of the mirror, what will be the characteristics of the image?

Answer 1

(a) Let's first find the image distance from the mirror equation:

$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f},$$ $$\dfrac{1}{2.5f}+\dfrac{1}{d_i}=\dfrac{1}{f},$$ $$d_i=\dfrac{1}{\dfrac{1}{f}-\dfrac{1}{2.5f}}=\dfrac{2.5f}{1.5}=\dfrac{2.5\cdot5\ cm}{1.5}=8.33\ cm.$$

The sign plus means that the image is real and located in front of the mirror.

Let's find the magnification of the mirror from magnification equation:

$$M=-\dfrac{d_i}{d_o}=-\dfrac{8.33 cm}{12.5\ cm}=-0.67.$$

The sign minus means that the image is inverted and diminished in size.

We can find the size of the image as follows:

$$M=\dfrac{h_i}{h_o},$$ $$h_i=Mh_o=-0.67\cdot5.5\ cm=-3.68\ cm.$$

The sign minus means that the image is inverted.

Finally, the image of the object is real and located in front of the mirror. Also, the the image is inverted and diminished in size (the size of the image is 3.68 cm).

(b) Let's first find the image distance from the mirror equation (this time the mirror is convex, therefore the focal length is negative):

$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=-\dfrac{1}{f},$$ $$\dfrac{1}{2.5f}+\dfrac{1}{d_i}=-\dfrac{1}{f},$$ $$d_i=\dfrac{1}{-\dfrac{1}{f}-\dfrac{1}{2.5f}}=\dfrac{2.5f}{-3.5}=-\dfrac{2.5\cdot5\ cm}{3.5}=-3.57\ cm.$$

The sign minus means that the image is virtual and located behind the mirror.

Let's find the magnification of the mirror from magnification equation:

$$M=-\dfrac{d_i}{d_o}=-\dfrac{-3.57 cm}{12.5\ cm}=+0.29.$$

The sign plus means that the image is upright but diminished in size.

We can find the size of the image as follows:

$$M=\dfrac{h_i}{h_o},$$ $$h_i=Mh_o=0.29\cdot5.5\ cm=1.59\ cm.$$

The sign plus means that the image is upright.

Finally, the image of the object is virtual and located behind the mirror. Also, the image is upright and diminished in size (the size of the image is 1.59 cm).