Mebs

1/29/2012

MEBS 6002 Lighting Engineering MSc BSE, Faculty of Engineering, HKU Lecture 2 Photometry ( (some materials extracted from handouts of Mr. K.F. Chan)

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Solid angle 

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From this definition, an , “ideal” lamp can only output 683 lumens per W energy input. That is the theoretically th ti ll peak.

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• However both the pearl lamp and the ‘white’ lamp have a much lower luminance than the clear because the light is coming from a much larger area. Projected • *Projected Area: The projected area or apparent area of a source is the flat, plain area ‘seen’ from the specified direction.

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I  I o cos  But L  Lo called matt reflecting surface.

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By Inverse Square Law, illuminance received by a plane normal to the direction of light at Point B = I  D 2 Where I is the luminous intensity along the direction concerned.

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Consider a hemispherical shell of radius R around a light emitting surface, area S, at the centre. If S emits Φ lumens, all will land on the shell surface. Consider a horizontal slice of ring on the hemisphere at an angle θ from the vertical.

Thickness of ring  Rd Area of ring  2 ( R sin  ) Rd  2R 2 sin d Illuminance received by ring  Lumens received by ring  I I

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R2 2R 2 sin d

R For a perfectly diffusing surface, I  I o cos 

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Lumens on ring  2I o sin  cos d

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Total lumens on hemisphere for a perfectly diffusing surface   cos 2  2   2I o sin  cos d  2I o   I o  4 0  0

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For F a reflecting surface, fl ti f   ES   surface reflectance ; E  illuminance received S  surface area   ES  I o ; Io E  Lo   S E For a matt surface, L  Lo  in all directions.

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Luminous exitance is flux emitted per unit area of light emitting source. ...