# Chapter 29 Photometry free study material by TEACHING CARE online tuition and coaching classes

Chapter 29 Photometry free study material by TEACHING CARE online tuition and coaching classes

The branch of optics that deals with the study and measurement of the light energy is called photometry.

# Important Definitions.

## (1) Solid angle

The area of a spherical surface subtends an angle at the centre of the sphere. This angle is called solid (w) .

• w= Area of DA

r 2

• It’s unit is
• Solid angle subtended by the whole sphere at it’s centre is 4p

The total energy radiated by a source per second is called radiant flux. It’s S.I. unit is Watt (W).

## (3) Luminous flux (f)

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 (h)

The Ratio of luminous flux and radiant flux is called luminous efficiency i.e. h = f .

R

 Light source Flux (lumen) Efficiency (lumen/watt) 40 W tungsten bulb 465 12 60 W tungsten bulb 835 14 500 W tungsten bulb 9950 20 30 W fluorescent tube 1500 50

## (5) Luminous Intensity (L)

In a given direction it is defined as luminous flux per unit solid angle i.e.

L = f

w

®    Light energy sec ´ solid angle

¾¾S.I¾. u¾n¾it ®

= candela(Cd)

Note : @The luminous intensity of a point source is given by :

## (6) Illuminance or intensity of illumination (I)

L =  f  Þ

4p

f = 4p ´ (L)

The luminous flux incident per unit area of a surface is called illuminance.

I = f

A

• Unit : I. unit –

Lumen

m 2

or Lux (lx)

CGS unit – Phot

Point source

I =    f   Þ I µ 1

1Phot = 104 Lux = 1Lumen

cm2

• Intensity of illumination at a distance r from

4pr 2             r 2

f              1

Note : @In case of a parallel beam of light

I µ r .

Line source

I = 2prl Þ I µ r

@    If a luminous flux of 1 lumen is falling on an area of 1m2

surface will be 1 Lux.

of a surface, then the illuminance of that

## (7)Difference between illuminance (intensity of illumination) and luminance (Brightness) of a surface

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.

• Illuminance of surface is given by :

I = L

r 2

• For a given source L = constant so law of

I µ 1

r 2

; This is called. Inverse square

# Lambert’s Cosine Law of Illuminance.

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 q with the normal to the surface, then

• Illuminance of the surface

I L cosq

r 2

• For a given light source and point of illumination (e. L and r = constant)

I µ cosq

this is called Lambert’s

cosine law of illuminance. Þ I

max = L

r 2

= Io

(atq = 0o )

• For a given source and plane of illuminance (e. L and h = constant)

cosq h

r

so I =

L cos 3 q

h2

or I = Lh

r 3

i.e.

I µ cos 3 q

or I µ 1

r 3

Note : @I varies with distance as

r 2

for isotropic point source, as æ 1 ö

 ç ÷

r

for line source and is independent of

è ø

r in case of parallel beam.

# Photometer and Principle of Photometry.

A photometer is a device used to compare the illuminance of two sources.

Two sources of luminous intensity

L1 , and

L2 are placed at distances

r1 and r2

from the screen so that their

flux are perpendicular to the screen. The distance r1 and r2

I1 = I 2 .

L         L            L         æ r ö 2

So    1 =    2 Þ    1 = ç 1 ÷

; This is called principle of photometry.

 r
 r
 L2
 1          2

2        2                è r2 ø

Note : @ R µ f µ L

so that

R1 = f1

R2       f2

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 40 watt fluorescent tube will be more than that of the filament bulb of same wattage.

Example: 1           If luminous efficiency of a lamp is 2 lumen/watt and its luminous intensity is 42 candela, then power of the lamp is                                                                                                                                           [AFMC 1998]

(a) 62 W                                   (b) 76 W                              (c) 138 W                                (d) 264 W

Solution: (d)            Luminous flux = 4p L = 4 ´ 3.14 ´ 42 = 528 Lumen

Power of lamp =

Luminous flux Luminous efficiency

= 528 = 264 W

2

Example: 2           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                                       (b) 40                           (c) 20                              (d)

40 ´ 104

Solution: (b)

I L cosq

r 2

Þ  L =

I ´ r 2

cosq

= 5 ´ 10 4 ´ 104 ´ 22

cos 60°

= 40 Candela

Screen

Example: 3           In a movie hall, the distance between the projector and the screen is increased by 1% illumination on the screen is                                                                                                                                           [CPMT 1990]

• Increased by 1% (b) Decreased by 1%     (c) Increased by 2%            (d) Decreased by 2%

Solution: (d)

I = L

Þ dI = – 2dr

(Q L = constant)                Þ

dI ´ 100 = – 2 ´ dr ´ 100

= -2 ´ 1 = -2%

r 2              I             r                                                            I                          r

Example: 4 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

• 100 sec (b) 25 sec                            (c) 50 sec                                (d) 75 sec

Solution: (c)            For equal fogging I 2

• t 2

= I1

´ t1 Þ

L2 ´ t

 r
 2
 2

2

= L1 ´ t    Þ

 r
 2
 1

1

16 ´ t 2

4

= 20 ´ 10 Þ

1

t 2 = 50Sec.

Example: 5           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) I 0

(b)

2 5 I 0

(c)

2I 0

(d)

5 5 I 0

Solution: (d)            The illuminance at B

I B   = L

12

………(i)

and illuminance at point C I

L cosq    L     ´ 1   Þ I       L

……… (ii)

C        (             (                        0       5

From equation (i) and (ii)

I B = 5       I 0

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] (a) 100 : 1                        (b) 104 : 1                     (c) 1 : 100                       (d) 1 : 104

1             Illuminance on slide

(Length of image on screen)2

æ 3.5 m ö 2

Solution: (b)

I µ             So,                                             =

= ç               ÷

= 104 : 1

r 2               Illuminance on screen

(Length of object on slide)2

è 35 mm ø

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

(b) 2 / 1                        (c) 17 / 13                       (d)   5 / 4

Solution: (a)            The illuminance at A is I A =   L         ´ cosq 1 = L ´    =     3L

The illuminance at B is  I B =

13

L       ´ cosq 2 = L ´

17

(13)3 / 2

=      3L (17)3 / 2

I A

I

æ 17 ö 3 / 2

= ç 13 ÷

B       è      ø