Another of the technicians in Fig. 31.11 is standing 2.46 m in front of the mirror. (a) Find th

Another of the technicians in Fig. 31.11 is standing 2.46 m...

Another of the technicians in Fig. $31.11$ is standing $2.46 \mathrm{~m}$ in front of the mirror. (a) Find the location of the technician's image, (b) determine its magnification, and (c) tell whether it's real or virtual, upright or inverted.

Key Concepts

Sign Conventions in Mirror Optics

Sign conventions provide a consistent method for assigning positive and negative values to distances in mirror problems. They are crucial when applying the mirror equation and determining magnification. Typically, for mirrors, distances measured in the direction of incoming light (object side) are considered positive, while those measured in the opposite direction (image side) are negative. These conventions help in correctly predicting the nature and properties of the image formed.

Image Characteristics (Real vs. Virtual and Upright vs. Inverted)

Images formed by mirrors can be categorized based on their properties. A real image is formed when light rays actually converge at a point, and it is typically inverted relative to the object. In contrast, a virtual image occurs when light rays appear to diverge from a point behind the mirror, resulting in an image that is upright. Understanding whether an image is real or virtual, as well as its orientation, is crucial in the analysis of mirror problems.

Mirror Equation

The mirror equation, 1/f = 1/d? + 1/d?, is a fundamental relationship in geometric optics for spherical mirrors. It relates the focal length (f) of the mirror to the object distance (d?) and the image distance (d?). This equation is essential for determining where an image is formed relative to the mirror and for analyzing the mirror’s behavior in various situations.

Magnification

Magnification in mirror optics is defined as m = -d?/d?, which relates the image distance to the object distance. It is a measure of how much larger or smaller the image is relative to the object, and its sign indicates the orientation of the image. A negative magnification signifies that the image is inverted relative to the object, while a positive magnification indicates an upright image.

Transcript

00:01 So problem 3130 is a variation of problem or example 311 in the textbook, where a technician stands in front of the hubble space telescope concave mirror with a focal length of 5 .52 meters.

00:17 In this problem, the technician stands 2 .46 meters away from the mirror, so that is what we call the object distance, and we need to try to find the image distance.

00:27 So we can plug it into the equation 1 over 8.

00:31 F equals 1 over s plus 1 over s prime or the one over the object distance plus one over the image distance.

00:38 Vocal length is 5 .52 meters.

00:42 Object distance is 2 .46 meters and image distance is unknown.

00:48 So rearranging that equation, the image distance, the inverse of the image distance is 1 over 5 .52 minus 1 over 2 .46.

00:59 And the reciprocal tool or key in your calculator is helpful.

01:04 It makes it nice and easy to calculate that the image distance is negative, 4 .43 meters.

01:11 So just a couple of quick rays will confirm that this is a negative or a virtual image.

01:18 If we draw a ray from the technician's head through the focal length, and then one that goes right to the principal axis bounces off at an equal angle.

01:29 And these are pretty parallel.

01:32 Let me back this up and just do it one more time to make it a little more exaggerated there.

01:39 These two angles should be equal, but you can see that these two rays will never connect, so they are diverging.

01:46 So just to back them up a little bit.

01:50 You can see, not being to scale, but you can see that my image would be over here...

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