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Ray optics. Every point on a luminous or illuminated surface produces light rays in all available directions.

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Presentation on theme: "Ray optics. Every point on a luminous or illuminated surface produces light rays in all available directions."— Presentation transcript:

1 Ray optics

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7 Every point on a luminous or illuminated surface produces light rays in all available directions

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16 The image in a plane mirror is called a virtual image, meaning that no light rays pass through the image. Rather light rays only appear to originate at the image.

17 The paraxial approximation: All rays are nearly parallel to the optic axis. I.e. all angles are small

18 Spherical mirrors

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21 The small angle approximations are valid whenever the diameter of the mirror is small compared to its radius of curvature. In this case, tan( 2 ) is approximately equal to 2, so that

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23 The Principal of Reciprocity: If a light ray follows a certain path through an optical system, an incident ray in the opposite direction will follow the reversed path.

24 A ray going through the focal point will emerge parallel to the optic axis.

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34 Problem: A concave mirror has a radius of curvature of 24cm. An object of height 8mm is placed 40cm from the mirror. Find the location and height of the image.

35 Solution: f = 12cm, 1/s i + 1/s o = 1/f => s i = f s o / (s o – f) => s i = (12cm) (40cm) / [ 40cm – 12cm] = 480 cm 2 / 28 cm = 17.1cm h i = M h o = - (s i / s o ) h o = - (17.1 / 40) (8 mm) = mm

36 Problem: A concave mirror has a radius of curvature of 24cm. An object of height 8mm is placed 18cm from the mirror. Find the location and height of the image. Solution: f = 12cm, s i = f s o / (s o – f) => s i = (12cm) (18cm) / [ 18cm – 12cm] = 216 cm 2 / 6 cm = 36 cm h i = M h o = - (s i / s o ) h o = - (36 / 18) (8 mm) = - 16 mm

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45 Problem: A concave mirror has a radius of curvature of 24cm. An object of height 8mm is placed 6cm from the mirror. Find the location and height of the image.

46 Problem: A concave mirror has a radius of curvature of 24cm. An object of height 8mm is placed 6cm from the mirror. Find the location and height of the image. Solution: f = 12cm, 1/s i + 1/s o = 1/f => s i = f s o / (s o – f) => s i = (12) (6cm) / [ 6cm – 12cm] = 48 cm 2 / -6 cm = - 8 cm h i = M h o = - (s i / s o ) h o = - (-8 / 6) (8 mm) = mm

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55 Whenever the paraxial approximation is valid,

56 s i > 0 => real image s i virtual image M > 0 => upright image M inverted image


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