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Geometric Optics. Image Formation by Spherical Mirrors. Spherical Mirrors : Are shaped like sections of a sphere. May reflect on the inside (concave) or outside (convex). Approximation : Rays coming from far away objects are approximately parallel.
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Image Formation by Spherical Mirrors • Spherical Mirrors: Are shaped like sections of a sphere. • May reflect on the inside (concave) or outside (convex).
Approximation: Rays coming from far away objects are approximately parallel. • Consider such parallel rays striking a concave mirror. The law • of reflection holds at each point where a ray strikes the mirror. • But, if a mirror has large curvature, • parallel rays striking it will not all • converge at exactly the same place. • This is called spherical aberration& it causes a blurred image. Reflected rays don’t converge to the same point.
The Principle Axisis the straight line perpendicular to the mirror surface at it’s center. If the incoming rays are parallel to the principle axis & if the mirror is small compared to it’s radius of curvature, r, so that the reflected ray makes a small angle with respect to the incident ray, then, reflected rays will converge with each other at approximately a single point, called The Focal PointF. Focal length f. Radius of curvature r The distance between the focal point & the mirror center The focal length,f.
Using simple geometry, and assuming that the angle θ is small, it can be shown that the focal length is half the radius of curvature. Spherical aberrationcan be avoided by using a parabolic reflector; these are more difficult & expensive to make, & so are used only when necessary, such as in research telescopes.
Image Formation by Spherical Mirrors • Ray diagrams are used to determine where an image will be. For mirrors, three key rays, each beginning on the object, are used: • A ray parallel to the axis; after reflection it passes through the focal point. • A ray through the focal point; after reflection it is parallel to the axis. • A ray perpendicular to the mirror; it reflects back on itself. • See the figures on the next slide!
Three key rays, each beginning on the object, are used: object Ray 1: Goes out from O' parallel to the axis & reflects through F. Ray 2: Goes through F & then reflects back parallel to the axis Ray 3: Chosen perpendicular to the mirror & so must reflect back on itself & go through the center of curvature C image l
The intersection of these 3 rays gives the position of the image of that point on the object. To get a full image, do the same with other points (2 points are enough for many purposes). This is called The Mirror Equation A relation between the object distance dO, the image distance di, & the mirror focal lengthf can be obtained from simple geometry. The result is: Since light passes through the image, the image in this case is called a Real Image.It can be seen on a piece of paper at pointI or on a photographic film there. Contrast:For a plane mirror, no light passes through the image, so it is called a Virtual Image. Note: In your book do & di are denoted as so & si
Again using geometry, the magnification m (the ratio of the image height, hito the object height hO) is found to be: The negative sign indicates that the image is inverted. This object is between the center of curvature and the focal point, and its image is larger, inverted, and real.
Example: Image in a concave mirror. A diamond ring of height hO = 1.50 cm is placed a distance dO =20.0 cm from a concave mirror with radius of curvature r = 30.0 cm. Calculate (a) the position of the image, and (b) its size. Solution:The focal lengthf = (½)r = 15.0 cm. (a)Mirror Equation gives: (1/di) = (1/f) - (1/dO) So, di=60.0 cm (b) The magnification is then: m = - (di/dO) = -3.0 Note 1.di> 0, so the image is real (on the same side as the object). 2.m < 0, so the image is inverted.
Conceptual Example:Reversible rays. If the object in this figure is placed where the image is, where will the new image be? Figure 32-16 goes here.
If an object is outside the center of curvature of a concave mirror, its image will be inverted, smaller, and real.
Example: Object closer to concave mirror. A 1.00-cm-high object is placed 10.0 cm from a concave mirror whose radius of curvature is 30.0 cm. (a) Draw a ray diagram to locate (approximately) the position of the image. (b) Find the position of the image & the magnification.
Convex Mirrors Exactly the same type of analysis can holds. It is found that the mirror equation still holds in this case, if the radius of curvature r & the focal length f are considered to be negative. Also, the image is always virtual, upright, and smaller than the object.
Problem Solving: Spherical Mirrors • Draw a ray diagram;the image is where the rays intersect. • Apply the mirror and magnification equations. • Sign conventions:If the object, the image, or the focal point is on the reflective side of the mirror, its distance is positive, and negative otherwise. The magnification is positive if image is upright. It is negative otherwise. • Check that your solution agrees with the ray diagram.
Example Convex rearview mirror An external rearview car mirror is convex with a radius of curvature r = 16.0 m. Find the location of the image and its magnification for an object dO = 10.0 m from the mirror.
