Tabela pomiarowa

	I	r	rx	Ix	Φ
L.p	[cd]	[mm]	[mm]	[cd]	[lm]
1	19	335 ±5,56	385±5,56	25,09 ± 0,14	315,19±1,75
2	19	337±5,56	383±5,56	24,54± 0,14	308,23±1,75
3	19	340±5,56	380±5,56	23,73± 0,14	298,09±1,75
4	19	343±5,56	377±5,56	22,95± 0,14	288,30±1,75
5	19	346±5,56	374±5,56	22,20± 0,14	278,83±1,75
6	19	345±5,56	375±5,56	22,45± 0,14	281,95±1,75
7	19	360±5,56	360±5,56	19,00± 0,14	238,64±1,75
8	19	363±5,56	357±5,56	18,38± 0,14	230,82±1,75
9	19	367±5,56	353±5,56	17,58± 0,14	220,78±1,75
10	19	370±5,56	350±5,56	17,00± 0,14	213,54±1,75
11	19	385±5,56	335±5,56	14,39± 0,14	180,68±1,75
12	19	395±5,56	325±5,56	12,86± 0,14	161,55±1,75
13	19	392±5,56	328±5,56	13,30± 0,14	167,08±1,75
14	19	389±5,56	331±5,56	13,76± 0,14	172,78±1,75
15	19	385±5,56	335±5,56	14,39± 0,14	180,68±1,75

1. Obliczyć światłość badanego źródła I_x oraz strumień świetlny Φ. Wyliczyć wartość średnią I_x, Φ.

$$\mathbf{I}_{\mathbf{x}}\mathbf{= I}\frac{\mathbf{r}_{\mathbf{x}}^{\mathbf{2}}}{\mathbf{r}^{\mathbf{2}}}$$

$$I_{x_{1}} = I\frac{r_{x_{1}}^{2}}{r_{1}^{2}} = 19\frac{{385}^{2}}{{335}^{2}} = 25,09\ cd$$

$$I_{x_{2}} = I\frac{r_{x_{2}}^{2}}{r_{2}^{2}} = 19\frac{{383}^{2}}{{337}^{2}} = 24,54\ cd$$

$$I_{x_{3}} = I\frac{r_{x_{3}}^{2}}{r_{3}^{2}} = 19\frac{{380}^{2}}{{340}^{2}} = 23,73\ cd$$

$$I_{x_{4}} = I\frac{r_{x_{4}}^{2}}{r_{4}^{2}} = 19\frac{{377}^{2}}{{343}^{2}} = 22,95\ cd$$

$$I_{x_{5}} = I\frac{r_{x_{5}}^{2}}{r_{5}^{2}} = 19\frac{{374}^{2}}{{346}^{2}} = 22,20\ cd$$

$$I_{x_{6}} = I\frac{r_{x_{6}}^{2}}{r_{6}^{2}} = 19\frac{{375}^{2}}{{345}^{2}} = 22,45\ cd$$

$$I_{x_{7}} = I\frac{r_{x_{7}}^{2}}{r_{7}^{2}} = 19\frac{{360}^{2}}{{360}^{2}} = \ 19\ cd$$

$$I_{x_{8}} = I\frac{r_{x_{8}}^{2}}{r_{8}^{2}} = 19\frac{{357}^{2}}{{363}^{2}} = \ 18,38\ cd$$

$$I_{x_{9}} = I\frac{r_{x_{9}}^{2}}{r_{9}^{2}} = 19\frac{{353}^{2}}{{367}^{2}} = \ 17,58\ cd$$

$$I_{x_{10}} = I\frac{r_{x_{10}}^{2}}{r_{10}^{2}} = 19\frac{{350}^{2}}{{370}^{2}} = \ 17\ cd$$

$$I_{x_{11}} = I\frac{r_{x_{11}}^{2}}{r_{11}^{2}} = 19\frac{{335}^{2}}{{385}^{2}} = \ 14,39\ \text{cd}$$

$$I_{x_{12}} = I\frac{r_{x_{12}}^{2}}{r_{12}^{2}} = 19\frac{{325}^{2}}{{395}^{2}} = \ 12,86\ cd$$

$$I_{x_{13}} = I\frac{r_{x_{13}}^{2}}{r_{13}^{2}} = 19\frac{{328}^{2}}{{392}^{2}} = \ 13,30\ cd$$

$$I_{x_{14}} = I\frac{r_{x_{14}}^{2}}{r_{14}^{2}} = 19\frac{{331}^{2}}{{389}^{2}} = \ 13,76\ cd$$

$$I_{x_{15}} = I\frac{r_{x_{15}}^{2}}{r_{15}^{2}} = 19\frac{{335}^{2}}{{385}^{2}} = \ 14,39\ cd$$

Φ_xn=∫₀^4πIdΩ=4πI_xn

Φ_x1 = 4πI_x1 = 4π • 25, 09 = 315, 19 lm

Φ_x2 = 4πI_x2 = 4π • 24, 54 = 308, 23 lm

Φ_x3 = 4πI_x3 = 4π • 23, 73 = 298, 09 lm

Φ_x4 = 4πI_x4 = 4π • 22, 95 = 288, 30 lm

Φ_x5 = 4πI_x5 = 4π • 22, 20 = 278, 83 lm

Φ_x6 = 4πI_x6 = 4π • 22, 45 = 281, 95 lm

Φ_x7 = 4πI_x7 = 4π • 19, 00 = 238, 64 lm

Φ_x8 = 4πI_x8 = 4π • 18, 38 = 230, 82 lm

Φ_x9 = 4πI_x9 = 4π • 17, 58 = 220, 78 lm

Φ_x10 = 4πI_x10 = 4π • 17, 00 = 213, 54 lm

Φ_x11 = 4πI_x11 = 4π • 14, 39 = 180, 68 lm

Φ_x12 = 4πI_x12 = 4π • 12, 86 = 161, 55 lm

Φ_x13 = 4πI_x13 = 4π • 13, 30 = 167, 08 lm

Φ_x14 = 4πI_x14 = 4π • 13, 76 = 172, 78 lm

Φ_x15 = 4πI_x15 = 4π • 14, 39 = 180, 68 lm

$$\mathbf{I}_{\mathbf{x}_{\mathbf{sr}}}\mathbf{=}\frac{\sum_{}^{}\mathbf{I}_{\mathbf{x}}}{\mathbf{15}}\mathbf{=}\frac{\mathbf{281,61}}{\mathbf{15}}\mathbf{= 18,78\ cd}$$

$$\mathbf{\Phi}_{\mathbf{sr}}\mathbf{=}\frac{\sum_{}^{}\mathbf{\Phi}}{\mathbf{15}}\mathbf{=}\frac{\mathbf{3537,14}}{\mathbf{15}}\mathbf{= 235,81\ lm}$$

1. Dla wielkości r, r_x wyliczyć niepewności standardowe metodą typu A.

$$u\left( r \right) = \sqrt{\frac{\sum_{i = 1}^{n}\left( r_{i} - \overset{\bar{}}{r} \right)^{2}}{n\left( n - 1 \right)}} = \sqrt{\frac{6508}{210}}\text{mm} = \sqrt{31}mm = 5,56\ mm$$

$$u\left( r_{x} \right) = \sqrt{\frac{\sum_{i = 1}^{n}\left( r_{\text{ix}} - \overset{\bar{}}{r_{x}} \right)}{n\left( n - 1 \right)}} = \ \sqrt{\frac{6508}{210}} = \sqrt{31} = 5,56\ mm$$

1. Obliczyć niepewność złożoną u(I_x).

$u\left( I_{x} \right) = \sqrt{\left( \frac{I_{x}}{\text{ΑR}}u\left( r \right) \right)^{2} + \left( \frac{I_{x}}{r_{x}}u\left( r_{x} \right) \right)^{2}} = \sqrt{\left( \frac{- 2Irr_{x}^{2}}{r^{4}}u(r) \right)^{2} + \left( \frac{{2Ir}_{x}r^{2}}{r^{4}}u\left( r_{x} \right) \right)^{2}} = \sqrt{\left( \frac{- 2 \bullet 19 \bullet 363,5 \bullet {356,5}^{2}}{{363,5}^{4}} \right)^{2} + \left( \frac{- 2 \bullet 19 \bullet 356,5 \bullet {363,5}^{2}}{{363,5}^{4}} \right)^{2}\text{\ \ }} = \sqrt{{0,1005}^{2} + {0,1025}^{2\ }} = \ \sqrt{0,0206\ } = 0,14\ cd$

1. Z prawa przenoszenia niepewności obliczyć niepewność strumienia świetlnego u(Φ)

$$u\left( \Phi \right) = \frac{\text{dΦ}}{\text{dI}_{x}}u\left( I_{x} \right) = 4\pi u\left( I_{x} \right) = 4 \bullet 3,14\ \bullet 0,14 = 1,75\ lm$$