Can you draw a ray diagram with the correct angles. when I tried drawing the angles, I got values that were different by a few degrees.. why do you think that happened?

Materials: Protractor, paper, pins, corkboard, mirror, plexiglass block, ruler. Task 1: Reection When light reects off a at, smooth, shiny surface, the light follows a path whose directions can be described by a simple equation. Your task is to verify that equation for a couple of different angles of the incoming light. Set as many sheets of paper on the corkboard as there are members in your group. Stand the mirror near one edge of the paper. Trace the bottom edge of the mirror on the paper. Stick a white pin in the board at the surface of the mirror, near the middle. Place a blue pin somewhere you can see its reection in the mirror. Rotate the board or move your eye around until reection of the blue pin is hidden behind the white pin. Finally, stick a red pin in the board between your eye and the white pin so that you see three pins lined up: the red pin, the white pin and the reflection of the blue pin. Mirror Reflection of Blue Pin ' Reflection of White Pin $Blue Pin R ed Pin What you have done is to nd the path that light bouncing off the blue pin takes when it happens to head for a particular point on the surface of the mirror. Label your pin positions so you know which is which, and then do the process again with the rst pin at a different angle to the mirror. Twice is enough. Take the paper off the board and use a ruler to draw lines showing the path from the blue pin to the mirror and the path from the mirror to the red pin. Measure the angles these paths make with a line that is perpendicular to the surface of the mirror. What is the relationship between the angles of the incoming and exiting light at the mirror surface? When you nish with your sheet of paper, staple it to the previous page. Mirror When a light ray is incoming from the blue pin A Reflection of Blue Pin IN towards the mirror as shown then it makes Reflection of White an angle with the normal shown equal to Pin White Pin D angle of incidence ( Li) $ forms the reflection Blue Pin Normal (N ) Red Pin as shown . S Seeing from AADP $ APOQ , we can observe that OP = Common 0Q = Ab ( by measurement ) AP = PQ (by measurement So, 4AOP = APOQ Theis ( AOP = /Pog = 90 - Li Now, Since / AON = (NOS ( Vertically opposite | Egh of Reflection . 0 "Angles ) thees, Li = LY Angle Angle of of reflection incidence with normal