 # → ℎ

## Presentation on theme: "→ ℎ "— Presentation transcript:

→ ℎ 𝑜 𝑠 = ℎ 𝑖 𝑠 ′ → 𝑠 ′ 𝑠 = ℎ 𝑖 ℎ 𝑜 𝑚 = 𝑠 ′ 𝑠 = ℎ 𝑖 ℎ 𝑜 𝜃 𝑖 = 𝜃 𝑟 =𝛼 →𝑡𝑎𝑛 𝜃 𝑖 =𝑡𝑎𝑛𝛼 𝑚=− 𝑠 ′ 𝑠 = ℎ 𝑖 ℎ 𝑜 The sign on the magnification is related to the orientation of the object and resulting image.

Spherical Mirror: Spherical mirrors are either concave or convex. Concave – Curved towards object – converging mirror Convex – Curved away from object – diverging mirror Concave mirror c f o c i f Principle Axis Focal point Center of Curvature When 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 image The focal point is located at a distance ½ the radius of curvature.

1 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 image The following can be determined from geometry. The Mirror Equation: Magnification: ho – Object height hi – Image height s – Object distance from mirror s’ – Image distance from mirror 1 𝑠 + 1 𝑠 ′ = 1 𝑓 𝑚=− 𝑠 ′ 𝑠 = ℎ 𝑖 ℎ 𝑜 Relates object distance, image distance and focal point.

Constructing 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. f 2) The incident ray passes through the focal point, and the reflected ray is parallel to the principle axis. i o c 3) 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 Mirrors Positive location – in front of mirror Negative location – behind mirror Positive height – upright Negative height – inverted Positive focal length – concave Negative focal length - convex o i c f

Example: 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) Convex S = 20 cm ho = 3 cm f = -10 cm 1 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)