Photometry Chapter 36 PDF

60 Photometry 177 E3 The branch of optics that deals with the study and measurement of the light energy is called photometry. Important Definitions. (1) Solid angle (i)   D YG U ID The area of a spherical surface subtends an angle at the centre of the sphere. This angle is called solid ( ). Area...

60 Photometry 177 E3 The branch of optics that deals with the study and measurement of the light energy is called photometry. Important Definitions. (1) Solid angle (i)   D YG U ID The area of a spherical surface subtends an angle at the centre of the sphere. This angle is called solid ( ). Area of A r2 (ii) It’s unit is steradian. (iii) Solid angle subtended by the whole sphere at it’s centre is 4 radians. (2) Radiant flux (R) U The total energy radiated by a source per second is called radiant flux. It’s S.I. unit is Watt (W). (3) Luminous flux () ST The total light energy emitted by a source per second is called luminous flux. It represents the total brightness producing capacity of the source. It’s S.I. unit is Lumen (lm). Note :  The luminous flux of a source of (1/685) watt emitting monochromatic light of wavelength 5500 Å is called 1 lumen. (4) Luminous efficiency () The Ratio of luminous flux and radiant flux is called luminous efficiency i.e.    R. Light source Flux (lumen) Efficiency (lumen/watt) 40 W tungsten bulb 465 12 178 Photometry 60 W tungsten bulb 835 14 500 W tungsten bulb 9950 20 30 W fluorescent tube 1500 50 60 (5) Luminous Intensity (L) In a given direction it is defined as luminous flux per unit solid angle i.e. L  Light energy lumen S.I. unit   candela (Cd ) sec  solid angle steradian  (6) Illuminance or intensity of illumination (I) L     4   ( L) 4 E3 Note :  The luminous intensity of a point source is given by : (i) Unit : S.I. unit – Lumen m2 or Lux (lx) 1 Phot  10 Lux  4 1 Lumen cm 2 U CGS unit – Phot ID The luminous flux incident per unit area of a surface is called illuminance. I  Point source I  A  1 I 2 2 4r r D YG (ii) Intensity of illumination at a distance r from Note :  In case of a parallel beam of light Line source I  1 I 2rl r I  r.  If a luminous flux of 1 lumen is falling on an area of 1m 2 of a surface, then the illuminance of that surface will be 1 Lux. U (7) Difference between illuminance (intensity of illumination) and luminance (Brightness) of a surface ST The illuminance represents the luminous flux incident on unit area of the surface, while luminance represents the luminous flux reflected from a unit area of the surface. Relation Between Luminous Intensity (L) and Illuminance (I). If S is a unidirectional point source of light of luminous intensity L and there is a surface at a distance r from source, on which light is falling normally. (1) Illuminance of surface is given by : I  L r2 (2) For a given source L = constant so I  Inverse square law of illuminance. Lambert’s Cosine Law of Illuminance. 1 r2 ; This is called. Photometry 179 In the above discussion if surface is so oriented that light from the source falls, on it obliquely and the central ray of light makes an angle  with the normal to the surface, then L cos  r2 60 (1) Illuminance of the surface I  (2) For a given light source and point of illumination (i.e. L and r = constant) I  cos  this is L S called Lambert’s cosine law of illuminance.  I max  2  Io (at   0 o ) r E3  (3) For a given source and plane of illuminance (i.e. L and h = constant)h L Lh 1 h so I  2 cos 3  or I  3 i.e. I  cos 3  or I  3 r r h r Note :  I varies with distance as 1 r 2 P0  P 1 for isotropic point source, as   for line source and r is independent of r in case of parallel beam. ID cos  r U Photometer and Principle of Photometry. D YG A photometer is a device used to compare the illuminance of two sources. L1 r1 L2 r2 U Two sources of luminous intensity L 1 , and L 2 are placed at distances r1 and r2 from the screen so that their flux are perpendicular to the screen. The distance r1 and r2 are adjusted till ST I1  I 2. 2 r  L So 2  2  1   1  ; This is called principle of photometry. L 2  r2  r1 r2 Note :  R    L so that R1  1  L1 R 2  2 L2 L1 L2  40 watt fluorescent tube gives more light than a filament bulb of same wattage because filament bulb emits light along with ultraviolet and infrared radiation. In a fluorescent tube, gas discharge produces only light and ultraviolet radiation. Since ultraviolet radiations too are converted into visible light through the phenomenon of photoluminescence, the illuminance, luminous flux or luminous efficiency of a 180 Photometry 40 watt fluorescent tube will be more than that of the filament bulb of same wattage. Example s (a) 62 W Luminous flux 528  Luminous efficiency 2 I (b) 40 L cos   L r2 I  r2 cos  I L 2  U Solution: (d) E3 Screen (d) Decreased by 2% dI 2  dr  100    100  2  1  2 % I r  Correct exposure for a photographic print is 10 seconds at a distance of one metre from a point source of 20 candela. For an equal fogging of the print placed at a distance of 2 m from a 16 candela source, the necessary time for exposure is (b) 25 sec For equal fogging I 2  t 2  I1  t1  ST Example: 5 2m (b) Decreased by 1% (c) Increased by 2% dI 2dr ( L = constant)  I r (a) 100 sec Solution: (c) 60o In a movie hall, the distance between the projector and the screen is increased by 1% illumination on the screen is r Example: 4 Normal 5  10 4  10 4  2 2  40 Candela cos 60  (a) Increased by 1% Solution: (d) (d) 40  10 4 (c) 20 D YG  Example: 3  264 W An electric bulb illuminates a plane surface. The intensity of illumination on the surface at a point 2m away from the bulb is 5  10 4 phot (lumen/cm2). The line joining the bulb to the point makes an angle of 60o with the normal to the surface. The intensity of the bulb in candela is [IIT-JEE 1980; CPMT 1991] (a) 40 3 Solution: (b) (d) 264 W Luminous flux  4 L  4  3.14  42  528 Lumen Power of lamp  Example: 2 (c) 138 W ID Solution: (d) (b) 76 W 60 If luminous efficiency of a lamp is 2 lumen/watt and its luminous intensity is 42 candela, then power of the lamp is U Example: 1 (c) 50 sec L2 r22 L1  t2  r12  t1  (d) 75 sec 16 20  t2   10  t 2  50 Sec. 4 1 A bulb of 100 watt is hanging at a height of one meter above the centre of a circular table of diameter 4 m. If the intensity at a point on its rim is I0 , then the intensity at the centre of the table will be [CPMT 1996] (a) I0 (b) 2 5 I0 The illuminance at B IB  (c) 2I0 L.........(i) A 12 and illuminance at point C I C  L cos  ( 5) From equation (i) and (ii) (d) 5 5 I0 2 I B  5 5 I0  L ( 5)  2 1 5  I0  L 5 5 Lamp  5m......... (ii)1 m B  2m Photometry 181 Example: 6 A movie projector forms an image 3.5m long of an object 35 mm. Supposing there is negligible absorption of light by aperture then illuminance on slide and screen will be in the ratio of [CPMT 1982] 4 (a) 100 : 1 (b) 10 : 1 (c) 1 : 100 (d) 1 : 104 1  3.5 m Illuminanc e on slide (Length of image on screen) 2  So,   2 Illuminanc e on screen (Length of object on slide)  35 mm 2    10 4 : 1  I Example: 7 A 60 watt bulb is hung over the center of a table 4'4' at a height of 3'. The ratio of the intensities of illumination at a point on the centre of the edge and on the corner of the table is [CPMT 1976, 84] (a) (17 / 13 )3 / 2 Solution: (a) (b) 2 / 1 The illuminance at A is I A  (c) 17 / 13 L ( 13 ) 2  cos  1  (d) 5 / 4 E3 r2 60 Solution: (b) L 3 3L   13 (13 ) 3 / 2 13 1 2 13m 3m 3/2 ST U D YG U I  17   A   I B  13  L 3 3L   cos  2    2 17 (17 ) 3 / 2 ( 17 ) 17 ID The illuminance at B is I B L 17 m 2 2 2m B A