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BASIC DEFINITIONS 2 PRINCIPLES OF PHOTOMETRY 3 LUMMER – BRODHUN PHOTOMETER 4 PROBLEMS

PH 0101 UNIT- 3 LECT - 1. BASIC DEFINITIONS 2 PRINCIPLES OF PHOTOMETRY 3 LUMMER – BRODHUN PHOTOMETER 4 PROBLEMS. PHOTOMETRY Introduction : The light emitted or reflected by the objects can be measured in comparison with a standard source.

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BASIC DEFINITIONS 2 PRINCIPLES OF PHOTOMETRY 3 LUMMER – BRODHUN PHOTOMETER 4 PROBLEMS

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  1. PH 0101 UNIT- 3 LECT - 1 • BASIC DEFINITIONS 2 PRINCIPLES OF PHOTOMETRY 3 LUMMER – BRODHUN PHOTOMETER 4 PROBLEMS UNIT III LECT - 1

  2. PHOTOMETRY • Introduction : • The light emitted or reflected by the objects can be measured in comparison with a standard source. • The branch of optics which deals with the measurement of light is known as ‘Photometry’. (‘Photo’ means light and ‘metry’ means measurement). UNIT III LECT - 1

  3. BASIC DEFINITION : • Luminous Flux (F) : • The amount of light energy radiated from a source or an illuminating object in all directions per second is known as ‘luminous flux’. It is denoted by the letter F or and its unit is ‘lumen’. • Definition of 1 Lumen : 1 Lumen = UNIT III LECT - 1

  4. Luminous Intensity or Illuminating Power (I) : The luminous intensity (or) illuminating power (I) of a point source in any direction is defined as the luminous flux emitted or radiated per unit solid angle in that direction. If ‘F’ is the luminous flux radiated by a source within a solid angle ‘’ in any particular direction, then luminous intensity I = F /  UNIT III LECT - 1

  5. ‘F’ is measured in lumen and ‘’ in steradian , the unit for ‘I’ Definition of One Candela : It is the luminous intensity of one sixtieth (1/60)th of a square centimetre of platinum at its fusion point. (1769oC) or one candela is the luminous intensity of the source if it emits 1 lumen per steradian. UNIT III LECT - 1

  6. Illumination (or) Intensity of Illumination (E) : The illumination (or) intensity of illumination at a point on the surface is defined as the luminous flux (F) received on an unit area (A) of the surface surrounding the point. It’s unit is lumen /m2. UNIT III LECT - 1

  7. Other units of illumination : Lux : Itis equal to 1 lumen per square metre. It can also be defined as the intensity of illumination over a surface area of 1m2, placed at a distance of 1m from a point source of 1 Candela. Phot :It is the intensity of illumination of a source which receives one lumen per square centimetre. 1 phot = 104 lux. UNIT III LECT - 1

  8. Inverse square law of illumination : The intensity of illumination (E) of a surface due to a light source is directly proportional to the luminous intensity or illuminating power (I) of the source and inversely proportional to the square of the distance between the source and the surface (r2). (or) UNIT III LECT - 1

  9. Principle of Photometry : The principle of photometry is based on the adjustment of the distances of the two light sources from the measuring point they produce equal intensity of illumination at the point. UNIT III LECT - 1

  10. E1 and E2 be the intensity of illumination due to two sources S1 and S2 at a surface ‘S’,r1 and r2 be the distances of the two sources from the surface. The inverse square law of illumination, and Where I1 and I 2 are the luminous intensities of the two sources. By adjusting r1 and r2 , E1 and E2 are made equal. i.e., E1 = E2 UNIT III LECT - 1

  11.  = (or) = The luminous intensity of one source is known, by comparison, the luminous intensity of another source can be determined. Photometer : It is an instrument which is used to compare the luminous intensities of two sources. If the luminous intensity of one of the source is known, the luminous intensity of the other one can be measured. UNIT III LECT - 1

  12. LUMMER – BRODHUN PHOTOMETER : This is one of the most accurate photometers used for comparing the luminous intensities of two sources. LUMMER – BRODHUN PHOTOMETER UNIT III LECT - 1

  13. Screen ‘S’ :    This is a specially prepared screen made of some white diffusing material of high reflecting power such as plaster of paris or magnesium carbonate.    The light rays from the two sources are incident on the two sides of this screen. UNIT III LECT - 1

  14. Totally reflecting prism P1 and P2 : P1 and P2 are two isosceles right angled prisms whose angles are 45o, 90o and 45 o. The light rays scattered from the screen ‘S’, fall normally on one face of the prism, enter into the prisms without any change in their direction, undergo total internal reflection and emerge normally from the other face of these prisms. UNIT III LECT - 1

  15. Lummer Brodhun Cube A and B : • It consists of two right-angled isosceles prisms A and B in contact with each other. •       The edges of prism A are cut in such a way that it is slightly curved at the outer edge while flat in the central part. •       The prisms are cemented together at the central part with Canada Balsam, whose refractive index is the same as that of the material of the prisms. UNIT III LECT - 1

  16. The two prisms are in optical contact with each other at the central part and enclose an air film between them at other parts. Telescope T : The telescope (T) is used to receive the light coming from Lummer – Brodhun cube. The whole apparatus is enclosed in a metallic box blackened from inside. UNIT III LECT - 1

  17. Working : • S1 and S2are two sources whose luminous intensities are to be compared. • These sources can be moved along an optical bench. • They are placed on the opposite side of the screen ‘S’ at equal distances and illuminate the screen normally. • The light rays reflected diffusively from each side of screen are incident on the prisms P1 and P2. • These light rays undergo total internal reflection from the prisms and fall on Lummer Brodhun cube. UNIT III LECT - 1

  18. Contd. • The outer rays from the source S1 undergo total internal reflection at prism ‘A’, while the middle ray passes into the prism ‘B’ through the point of contact of the prisms A and B without any deviation. • The outer rays from S2 undergo total internal reflection at ‘B’, while the middle ray passes into the prism ‘A’ through the point of contact without any deviation. UNIT III LECT - 1

  19. Contd. • In this way, middle ray from source S1 and outer rays from source S2 enter into the telescope. • Since the luminous intensities of the two sources are different, the field of view of the eyepiece of the telescope consists of a central part illuminated by light from S1 and an outer part illuminated by light from S2 of different brightness. • Now the distances of S1 and S2are adjusted in such a way that the field of view appears equally bright. UNIT III LECT - 1

  20. Field of view in a Lummer – Brodhun Photometer When the two sources are of different intensities When the sources are having equal intensities. UNIT III LECT - 1

  21. Comparison of luminous intensities of two sources : When the distances of S1 and S2 are so adjusted that the illuminations on the two sides of ‘S’ are equal, the whole field of view appears uniformly bright and the boundary dividing the two fields disappears. Let I1 and I 2 be the luminous intensities of S1 and S2 and r1and r2 are their respective distances from ‘S’ , then according to inverse square law of illumination, UNIT III LECT - 1

  22. (or) = The luminous intensities of the two sources can be compared. If ‘I1’ is known, ‘I2’ can be calculated from the relation, (or) Thus the luminous intensities of the two sources can be compared. UNIT III LECT - 1

  23. This photometer is best suited for the comparison of luminous intensities of two light sources producing same colour. If the sources produce different colour the readings are unreliable, and flicker photometer is used. UNIT III LECT - 1

  24. Solved Problem (1) : Two lamps of 32 candle power and 8 candle power respectively are placed 2 metres apart. A screen is placed between them. Where should the screen be placed in order that it may be equally illuminated by the lamps on both sides ? Given I1 = 32 candle power I2 = 8 candle power r1+ r2 = 2m UNIT III LECT - 1

  25. Formula : Let ‘x’ be the distance from where the screen has to be placed to get equal illumination on the screen. Then r1 = x UNIT III LECT - 1

  26. 32(4+x2-4x) = 8x2 24x2 – 128x + 128 = 0 3x2 – 16x + 16 = 0 3x2 – 4x – 12x + 16 = 0 r1(3x-4)-4(3x-4) = 0 (x-4) (3x-4) = 0 Hence x can be either 4 or 4/3 Since the separation between the two lamps in 2m, x cannot be 4 m. Here x = 4/3m. i.e. when the screen is placed at a distance of 4/3m from the lamp of 32 candle power, the screen is illuminated equally by the lamps on both sides. UNIT III LECT - 1

  27. Solved Problem (2) : Two lamps of 32.c.p. and 8 c.p. respectively are placed 2 meters apart. Where should a screen be placed in order that it may be equally illuminated by the lamps ? There are two possible positions of the screen. (i) Both the sources lie on the same side of the screen.Let the distance of the first lamp (8 c.p.) from the screen be x as shown. When the screen is equally illuminated by the two lamps, then UNIT III LECT - 1

  28. UNIT III LECT - 1

  29. or   x = 200 cm. Thus the screen should be placed at a distance 200 cm from the lamp L1(8 c.p.) and 400 cm from the second lamp L2 (32 c.p.) UNIT III LECT - 1

  30. (ii) The screen is between two sources. When the screen is equally illuminated by the two lamps.. x = 66.67 cm. Thus the screen should be placed at a distance of 66.67 cm from the first lamp L1 (8 c.p.) and 133.33 cm from the second lamp L2 (32 c.p.) UNIT III LECT - 1

  31. Solved Problem (3) : Two identical sources are at a distance of 60 cm from each other. Where must a screen be placed in between them so that one side of the screen may be illuminated four times greater than the other  Let the screen be placed at a distance xcm from first source. Now the distance of the second source will be (60-x) cm. If the illumination of one side is E, then the illumination of the other side should be 4E. Let both the sources have illuminating power I. Then UNIT III LECT - 1

  32. Case. (i) Considering +ve sign 120-2x = x or x = 40 cm. Case. (ii) Considering –ve sign 120 – 2x = - x or x = 120 cm. which is impossible because it is greater than 60cm. Thus the distance of the screen must be 40 cm from one source and 20cm from the other source. UNIT III LECT - 1

  33. Exercise Problem (1) : Two lamps of 20 candle power and 10 candle power are placed A metre apart. A screen is placed such that both the lamps are on the same side of the screen and line joining the lamps is perpendicular to the screen. At what distance the screen has to be placed so that it may be equally illuminated by the lamps placed on the same side. At a distance of 9.655m from the lamp of 10 candle power. UNIT III LECT - 1

  34. Thank u UNIT III LECT - 1

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