It is important to note that the light is not separated into its component colors because it is not being “bent” or refracted, and all wavelengths are being reflected at equal angles. This concept is often termed the Law of Reflection. Thus, the angle of incidence is equal to the angle of reflection for visible light as well as for all other wavelengths of the electromagnetic radiation spectrum. Visible white light that is directed onto the surface of a mirror at an angle (incident) is reflected back into space by the mirror surface at another angle (reflected) that is equal to the incident angle, as presented for the action of a beam of light from a flashlight on a smooth, flat mirror in Figure 2. The incoming light wave is referred to as an incident wave, and the wave that is bounced away from the surface is termed the reflected wave. However, it wasn’t until a millennium and a half later that the Arab scientist Alhazen proposed a law describing exactly what happens to a light ray when it strikes a smooth surface and then bounces off into space. (c) Height of image A'B': 0.7 × 5 = 3.5 cm, i.e., image is smaller than the object.Some of the earliest accounts of light reflection originate from the ancient Greek mathematician Euclid, who conducted a series of experiments around 300 BC, and appears to have had a good understanding of how light is reflected. (b) Nature of image A’B’: Real and inverted. (a) Position of image A'B' = 3.3 cm × 5 = 16.5 cm from the lens on opposite side. (xi) Thus the final position, nature and size of the image A'B' are: It is found that CB' = 3.3 cm and A'B' = 0.7 cm. (ix) Now AB', represents the real, but inverted image of the object AB. (viii) Draw AB', perpendicular to the principal axis from A'. (vii) Let the two lines starting from A meet at A'. (vi) Draw a line from A to C (centre of the lens), which goes straight without deviation. (v) Draw a line AD parallel to principal axis and then, allow it to pass straight through the focus (F') on the right side of the lens. (iv) Draw an arrow AB of height 1 cm on the left side of lens at a distance of 5 cm from the lens. (iii) Mark two foci F and F' on two sides of the lens, each at a distance of 2 cm from the lens. (i) Draw a horizontal line to represent the principal axis of the convex lens. ![]() Therefore, on this scale 5 cm high object, object distance of 25 cm and focal length of 10 cm can be represented by 1 cm high, 5 cm and 2 cm lines respectively. As the distances given in the question are large, so we choose a scale of 1: 5, i.e., 1 cm represents 5 cm. Thus, object is placed at a distance of 6 cm × 5 = 30 cm from the lens.Ĭonverging lens means a convex lens. (viii) Draw a line AB, perpendicular (downwards) from A to meet the principal axis at B. (vii) Draw a line CA', backwards, so that it meets the line from D parallel to principal axis at A. (vi) Draw a line A'B', perpendicular to principal axis from B'. (v) Draw a line AD, parallel to principal axis. (iv) Join any point D (nearly at the top of lens) and F by a dotted line. (iii) Mark points F and B on the left side of lens at a distance of 3 cm and 2 cm respectively. (ii) Draw a convex lens, keeping principal centre (C) on the principal axis. (i) Draw the principal axis (a horizontal line). Image distance, v = - 10 cm ĭrawing the ray diagram: Using a scale of 1: 5, we get v = - 2 cm, f = - 3 cm.
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