Presentation on theme: "→ ℎ "— Presentation transcript:
1→ ℎ 𝑜 𝑠 = ℎ 𝑖 𝑠 ′→ 𝑠 ′ 𝑠 = ℎ 𝑖 ℎ 𝑜𝑚 = 𝑠 ′ 𝑠 = ℎ 𝑖 ℎ 𝑜𝜃 𝑖 = 𝜃 𝑟 =𝛼→𝑡𝑎𝑛 𝜃 𝑖 =𝑡𝑎𝑛𝛼𝑚=− 𝑠 ′ 𝑠 = ℎ 𝑖 ℎ 𝑜The sign on the magnification is related to the orientation of the object and resulting image.
2Spherical Mirror:Spherical mirrors are either concave or convex.Concave – Curved towards object – converging mirrorConvex – Curved away from object – diverging mirrorConcave mirrorcfocifPrinciple AxisFocal pointCenter of CurvatureWhen parallel light rays are incident on the concave mirror, it redirects them all to a single point called the focal point.Light rays from an object will converge at the location of the image.The image will be a real imageThe focal point is located at a distance ½ the radius of curvature.
31 𝑠 + 1 𝑠 ′ = 1 𝑓 Convex mirror f c o c i f Light rays from an object will diverge from the location of the image.When parallel light rays are incident on the convex mirror, it redirects them all away from the focal point.The image will be a virtual imageThe following can be determined from geometry.The Mirror Equation:Magnification:ho – Object heighthi – Image heights – Object distance from mirrors’ – Image distance from mirror1 𝑠 + 1 𝑠 ′ = 1 𝑓𝑚=− 𝑠 ′ 𝑠 = ℎ 𝑖 ℎ 𝑜Relates object distance, image distance and focal point.
4Constructing Mirror Ray Diagrams: Draw 3 rays from top of object.1) The Incident ray is drawn parallel to principle axis, and the reflected ray passes through the focal point.f2) The incident ray passes through the focal point, and the reflected ray is parallel to the principle axis.ioc3) The incident and reflected rays are drawn along the same line passing through the center of curvature.The image is located at the intersection of these three rays.Sign Convention for MirrorsPositive location – in front of mirrorNegative location – behind mirrorPositive height – uprightNegative height – invertedPositive focal length – concaveNegative focal length - convexoicf
5Example: An object 3 cm high is placed 20 cm from (a) a convex and (b) a concave spherical mirror, each of 10 cm focal length. Determine the position and nature of the image in each case.a) ConvexS = 20 cmho = 3 cmf = -10 cm1 𝑠 + 1 𝑠 ′ = 1 𝑓→ 1 𝑠 ′ = 1 𝑓 − 1 𝑠=− 1 10 𝑐𝑚 − 1 20 𝑐𝑚→ 1 𝑠 ′ = 3 20 𝑐𝑚→ 𝑠 ′ =−6.67 cm→ 𝑠 ′ =6.67 cm behind mirror (virtual image)