Photometry and Luminous Intensity - Class VIII

For same illumination $\dfrac{Power} { r^{2}}$ will be same

Luminous flux of a 10W source=Relative luminosity$\times$ luminous flux of 555nm source of 10W

Since $\dfrac {dF}{dA}$ is a constant, where $F$ is the illuminance and $A$ is the area of cross section. Hence the illuminance produced 3 m away does not change and it is 60 lux.

Case(i)
$\displaystyle\ \frac{I _{1}}{I _{2}}$ = $\displaystyle\ \left (\frac{80}{20} \right)^{2}$

The luminosity increases with increase in wavelenth, reaches a peak at around 550 nm then decreases.

The parameter that determines the brightness of a light source sensed by an eye is total luminous flux entering the eye.
$B = F$ where $B$ is the brightness and $F$ is the total flux entering the eye.

$E$  $\propto\displaystyle\ \frac{1}{r^{2}}$

Illuminance = $\dfrac{Luminous \ intensity  \times 4\pi } {area}$

$\alpha$ = $\displaystyle\ \frac{I _{2}}{I _{2}}$ = $\displaystyle\ \left(\frac{60}{75} \right)^{2}$ = $\displaystyle\ \left( \frac{4}{5} \right)^{2}$ = $\displaystyle\ \frac{16}{25}$ = $0.64$

Intensity of illumination = $\dfrac {L} {d^{2}}$

Illuminance ($E$)  is  measured  in  lux.  The  lux  is  an  SI  unit  used  when  characterizing illumination conditions of a surface:
$E = \dfrac {I}{R^2}    lux$
where $I$ is the luminous intensity and $R$ is the distance.
Given $I = 300 cd$ and $R = 10 m$
$\implies E = \dfrac {300}{10^2} = 3   lux$

Illuminance ($E$)  is  measured  in  lux.  The  lux  is  an  SI  unit  used  when characterizing illumination conditions of a surface:
$E = \dfrac {I}{R^2}     lux$
where $I$ is the luminous intensity and $R$ is the distance.
Given $I = 500 cd$ and $R = 10 m$
$\implies E = \dfrac {500}{10^2} = 5    lux$

$Illuminance = illuminating \ power * Cos \theta /d^{2}$

For the same impression $\dfrac{L * time}{d^{2}}$ has to be constant

Let distance between 8 Cd lamp and screen be $ x  m$

Illuminance ($E$)  is  measured  in  lux.  The  lux  is  an  SI  unit 

used  when  characterizing illumination conditions of a surface:
$E = \dfrac {I}{R^2}    lux$
$\implies R = \sqrt {\dfrac {I}{E}}$
where $I$ is the luminous intensity and $R$ is the distance.
Given $I = 50 cd$ and $E = 2 lm  m^{-2}$
$\implies R = \sqrt {\dfrac {50}{2}} = 5   m$

$ luminous \ flux = luminous \ intensity \times solid \ angle$

radiant flux in lumens = $\dfrac{3}{1.5 \times 10^{-3}} = 2000  lm$

We have total luminous flux,
$L = 4 \pi I = 4 \pi \times 100 = 400 \pi$

Inverse square law for illuminance is valid and can be applied for all types of sources.

The source of luminance of the blotting paper is the light reflected by it. Since 75 % of light is reflected,

$Luminous \ flux = luminous \ intensity * 4 \pi$

Illuminance is proportional to  Cos of angle at which light strikes the surface

We have illuminance
$I = \dfrac {\phi}{R^2}$
$\implies \phi = I  R^2 = 25 \times 3^2 = 225 \ Cd$

1 Candle power = 0.981 Candela

$ I = \dfrac {L Cos \theta}{r^\circ}$

Illuminance E at the surface, distance x away is $E= I/x^2$
here as the bulb is h distance away from the bulb. $E=I/h^2$

We have illuminance
$I = \dfrac {\phi}{R^2}$
$\implies \phi = I  R^2 = 10 \times 10^2 = 10^3  Cd$

We have illuminance,
$I \propto \dfrac {1}{R^2}$
where $R$ is the distance 
If the distance is doubled then 
$I' \propto \dfrac {I}{4}$

Illuminance depends on inverse of area and area is proportional to $r^{2}$ where is r is distance from the source to the point .
$illuminance \propto \dfrac{1}{r^{2}}$ 
$\dfrac{\partial (illuminance)}{\partial x}= -\dfrac{2\delta r}{r}$
hence changing distance by $1$% the illuminance will differ by $-\dfrac{2\delta r\times100}{r}= -2%$%
i.e,illuminance will decrease by $2$% 
option $C$ is correct

$Illuminance =\dfrac{ Luminous \ Power} { Area}$

The brightness producing capacity of a source "depends on its power" and "depends on the wavelength emitted" .
option $C,D$ are correct 

Luminous intensity, L = 20 Cd

Let intensity of incident radiation $ = I _{0}$
This incident radiation passes for 12s 
This implies the exposure is $ 12I _{0}$
If the photographic plate is at an angle $\theta$
Radiation passing through is given by:
$I = I _{0} cos\theta$ Given that $\theta = 60^{\circ} C$
$I = \dfrac{I _{0}}{2}$
For the same exposure:
$12I _{0} = \dfrac{I _{0}}{2} \times t$
i.e. we get $t = 24s$
Option C is correct.

Let $ I $ be the intensity of bulb then

The intensity is the power delivered per unit area, and hence it is inversely proportional to the square of the distance from source.

With increase in wavelength of visible light, the sensitivity to eye increases, reaches a peak at 555 nm and then decreases. So luminosity increases then decreases.

Only light of visible range excites the retina of eyes. The range Light, which can human eyes sense is called Luminus of light.
Hence, (b) is correct.

Luminous efficiency is the ratio of luminous flux to power. A standard 100% efficient source (monochromatic green light at 555 nm) produces 685 lumens/watt. Efficiency = (68.5 / 685) * 100% = 10%.

Illuminance E = (I / d^2) * cos(theta). Initially, E = I / 8^2 = I / 64. Finally, E = (I / 4^2) * cos(theta) = (I / 16) * cos(theta). Equating them: I / 64 = I / 16 * cos(theta) => cos(theta) = 16 / 64 = 1/4. Thus, theta = cos^-1(1/4).

Luminous flux $=\dfrac{light\ energy}{time}=\dfrac{10}{2}=5$ lumen

$F=4\pi l$
$F=(4\times 3.14)\times 100=1256$ lumen
$\therefore$ Luminous efficiency $=\dfrac{luminous\ flux}{electric\ power}=\dfrac{1256}{100}=12.56$

Distance from bulb to corner r = sqrt(2^2 + 1^2 + 1^2) = sqrt(6). Cos(theta) = height / r = 2 / sqrt(6). Illuminance E = (I / r^2) * cos(theta) = (60 / 6) * (2 / sqrt(6)) = 10 * 2 / 2.449 = 20 / 2.449 = 8.16 cd/m^2.

Luminous flux = luminous intensity * solid angle = $ 10* 4\pi = 400\pi$ lumen

At a distance $r$ from a line source of power$P$ and length $l$, the intensity will be

$E _2 = 4 E _1$. If x is distance from 1st source,
then, $\displaystyle \frac{I}{(1.2 - x)^2} = \frac{4 I}{x^2} $ or $\displaystyle \frac{1}{1.2 - x} = \frac{2}{x}$
$3x = 2.4, x = 0.8 m$

Illuminance ($E$)  is  measured  in  lux.  The  lux  is  an  SI  unit 

used  when  characterizing illumination conditions of a surface:
$E = \frac {I}{R^2}  cos \alpha     lux$
where $I$ is the luminous intensity, $R$ is the distance and $\alpha$ is the surface angle.
Given $I = 300 cd$, $R = 10 m$, $\alpha = 60^o$
$\implies E = \frac {300}{10^2}    cos  60^o= 1.5   lux$

$E _2 = \displaystyle \frac{1}{8} E _1$ or $\displaystyle \frac{1}{(r^2 + h^2)} \times \frac{h}{\sqrt{r^2 + h^2}} = \frac{1}{8} \frac{1}{h^2}$
(by lambert's cosine law)
or, $(r^2 + h^2)^{3/2} = (2h)^3 $ or $(r^2 + h^2)^{1/2} = 2h$
or $r^2 + h^2 = 4h^2$
$h = r / \sqrt{3}$