A spherical lens or mirror surface has a center of curvature located either along or decentered from the system local optical axis. {\displaystyle r=0} 0 The radius of curvature of a convex mirror used for rearview on a car is 4.00 m. If the location of the bus is 6 meters from this mirror, find the position of the image formed. °ÆA!°Š"°Š#� $� %° °Å°S�Ä z , This article is about optical applications. K Optical surfaces with non-spherical profiles, such as the surfaces of aspheric lenses, also have a radius of curvature. A surface that converges a wavefront is taken a positive e.g. from the axis. K Sometimes this image is real (you can form it on a surface such as a piece of paper) and sometimes virtual (it cannot be formed on a surface) Curvature and the sign convention The radius of curvature (R) of a surface is taken as a positive number if it increases the curvature of the waves – i.e. 2 Looking at the diagram you can see that waves with no curvature have been converged to a point a distance f (the focal length) from the lens. α The vertex of the lens surface is located on the local optical axis. ĞÏà¡±á > şÿ y { şÿÿÿ x ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿì¥Á 7 ğ¿ bjbjUU .. 7| 7| ñ ÿÿ ÿÿ ÿÿ l H H H H , ¼ ¼ ¼ 8 ô 4 , *› ì H H " j j j -� -� -� …š ‡š ‡š ‡š ‡š ‡š ‡š$ œ 6� h «š 9 -� W� Ö -� -� -� «š K’ H H j j íy äš K’ K’ K’ -� L H x j j …š K’ -� …š K’ : K’ …š À X …š j °ıŞ“ñNÊ, � ¼ y� Ğ …š …š úš 0 *› …š �� I� �� …š K’ , , H H H H Ù Lenses Lenses affect the light that passes through them making an image of the object from which the light waves come. Waves and curved surfaces When light waves fall on a curved surface that surface changes the curvature of the wave. Sign conventions 4 • Light travels from left to right • A radius of curvature is positive if the surface is convex towards the left • Longitudinal distances are positive if pointing to the right • Lateral distances are positive if pointing up • Ray angles are positive if the ray direction is obtained by rotating the Solution: Given: The radius of curvature (R)= +4.00 m. Object distance(u) = -6.00 m. Image distance(v) = ? is the conic constant, as measured at the vertex (where and Formula used: $$f=\frac{R}{2}$$ $$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$ converges the waves and negative if it decreases the curvature of the ways – i.e. For each R, the convention is such that we make R > 0 if the light hits the curved surface before the center of curvature, and R < 0 if the opposite is true.