 # Photometry

## Concept Of Physics

### H C Verma

1   A source emits $45$ joules of energy in $15 s$. What is the radiant flux of the source ?

##### Solution :

Radiant Flux = $\frac {Tota \ energy \ emitted}{Time} = \frac{45}{15s} =3w$

2   A photographic plate records sufficiently intense lines when it is exposed for $12 s$ to a source of $10 W$. How long should it be exposed to a $12 W$ source radiating the light of same colour to get equally intense lines ?

##### Solution :

To get equally intense lines on the photographic plate, the radiant flux (energy) should be same. So, $10W \times 12sec = 12W \times t$ $\\$ $\Rightarrow t = \frac{10W \times 12sec}{12W} = 10sec$.

3   Using figure $(22.1)$, find the relative luminosity of wavelength (a) $480 nm$, (b) $520 nm$ (c) $580 nm$ and (d) $600 nm$.

##### Solution :

It can be found out from the graph by the student.

4   The relative luminosity of wavelength $600 nm$ is $0.6$. Find the radiant flux of $600 nm$ needed to produce the same brightness sensation as produced by $120 W$ of radiant flux at $555 nm$.

##### Solution :

Relative luminousity = $\frac {Luminous \ flux \ of \ a \ source \ of \ given \ wave \ length}{Luminous \ flux \ of \ a source \ of \ 555 nm \ of \ same \ power}$ $\\$ Let the radiant flux needed be P watt. $\\$ $A_o, 0.6 = \frac{Luminous \ flux \ of \ source \ 'P' \ watt }{685 \ p}$ $\\$ $\therefore$ Luminous flux of the source = $(685 P) \times 0.6 = 120 \times 685$ $\\$ $\Rightarrow = \frac{120}{0.6} = 200W$

5   The luminous flux of a monochromatic source of $1 W$ is $450 lumen/watt$. Find the relative luminosity at the wavelength emitted.

##### Solution :

The luminous flux of the given source of 1W is 450 lumen/watt $\\$ $\Rightarrow$ Relative luminosity = $\frac{Luminous \ flux \ of \ the \ source \ of \ given \ wavelength}{Luminous \ flux \ of \ 555 nm \ source \ of \ same \ power} = \frac{450}{685} = 66\%$ $\\$ [$\therefore$ Since, luminous flux of 555nm source of 1w = 685 lumen]

6   A source emits light of wavelengths $555 nm$ and $600 nm$. The radiant flux of the $555 nm$ part is 40 W and of the $600 nm$ part is $30 W$. The relative luminosity at $600 nm$ is 0.6. Find (a) the total radiant flux, (b) the total luminous flux, (c) the luminous efficiency.

##### Solution :

7   A source emits light of wavelengths $555 nm$ and $600 nm$. The radiant flux of the $555 nm$ part is 40 W and of the $600 nm$ part is $30 W$. The relative luminosity at $600 nm$ is 0.6. Find (a) the total radiant flux, (b) the total luminous flux, (c) the luminous efficiency.

##### Solution :

The radiant flux of 555nm part is 40W and of the 600nm part is 30W $\\$ a) Total radiant flux = 40W + 30W = 70W $\\$ b) Luminous flux = $(L.Flux)_{555nm} + (L.Fllux)_{600nm}$ $\\$ =$1 \times 40 \times 685 + 0.6 \times 30 \times 685 =39730$ lumen $\\$ c) Luminous efficiency = $\frac{Total \ luminous \ flux}{Total \ radiant \ flux} = \frac{39730}{70} = 567.6 lumen/W$

8   A light source emits monochromatic light of wavelength $555 nm$. The source consumes $100 W$ of electric power and emits $35 W$ of radiant flux. Calculate the overall luminous efficiency.

##### Solution :

Overall luminous efficiency = $\frac{Total \ luminous \ flux}{Power \ input} = \frac{35 \times 685}{100}$ = 239.75 lumen/W

9   A source emits $31.4 W$ of radiant flux distributed uniformly in all directions. The luminous efficiency is $60 lumen/watt$. What is the luminous intensity of the source ?

##### Solution :

Radiant flux = 31.4W, Sold angle = 4$\pi$ $\\$ Luminous efficiency = $60$ lumen/W $\\$ So, Luminous flux = 60 $\times$ 31.4 lumen $\\$ And luminous intensity = $\frac{Luminous \ Flux}{4 \pi} = \frac{60 \times 31.4}{4 \pi}=150$ candela

10   A point source emitting $628$ lumen of luminous flux uniformly in all directions is placed at the origin. Calculate the illuminance on a small area placed at $(1'0 m, 0, 0)$ in such a way that the normal to the area makes an angle of $37°$ with the X-axis

##### Solution : l = luminous intensity = $\frac{628}{4 \pi}$ = 50 Candela $\\$ r = 1m, $\theta$ = 37° $\\$ So, illuminance, E = $\frac {lcos \theta}{r^{2}} = \frac{ 50 \times cos 37°}{1^{2}}$ = 40 lux

11   The illuminance of a small area changes from $900 lumen/m^2$ to $400 lumen/m^2$ when it is shifted along its normal by $10 cm$. Assuming that it is illuminated by a point source placed on the normal, find the distance between the source and the area in the original position.

##### Solution : Let, l = Luminous intensity of source $\\$ $E_A = 900 lumen/m^{2}$ $\\$ $E_B = 400 lumen/m^{2}$ $\\$ Now, $E_a = \frac {lcos \theta}{X^{2}}$ and $E_B = \frac{lcos \theta}{(X +10)^{2})}$ $\\$ So, l = $\frac{E_AX^{2}}{cos \theta} = \frac{E_B(X + 10)^{2}}{cos \theta}$ $\\$ $\Rightarrow 900x^{2} = 400(x + 10)^{2} \Rightarrow \frac{X}{X + 10} = \frac{2}{3} \Rightarrow 3X = 2X + 20 \Rightarrow X =20$cm

12   A point source emitting light uniformly in all directions is placed $60 cm$ above a table-top. The illuminance at a point on the table-top, directly below the source, is $15 lux$. Find the illuminance at a point on the table-top $80 cm$ away from the first point.

##### Solution : Given that, $E_a$ = 15 lux $\frac{l_{0}}{60^{2}}$ $\\$ $\Rightarrow l_{0} = 15 \times (0.6)^{2} = 5.4$ candela $\\$ So, $E_B = \frac {l_{0}cos \theta}{(OB)^{2}} = \frac {5.4 \times (\frac{3}{5})}{1^{2}} = 3.24$ lux

13   Light from a point source falls on a small area placed perpendicular to the incident light. If the area is rotated about the incident light by an angle of 60°, by what fraction will the illuminance change ?

##### Solution :

14   Light from a point source falls on a small area placed perpendicular to the incident light. If the area is rotated about the incident light by an angle of $60°$, by what fraction will the illuminance change ?

##### Solution :

The illuminance will not change.

15   A student is studying a book placed near the edge of a circular table of radius R. A point source of light is suspended directly above the centre of the table. What should be the height of the source above the table so as to produce maximum illuminance at the position of the book ?

##### Solution : Let the height of source is 'h' and the luminous intensity in the normal direction is $l_0$, $\\$ So, illuminance at the book is given by, $\\$ E = $\frac{l_0 cos \theta} {r^2} = \frac {l_{0}h}{r^{3}} = \frac{l_{0}h}{( r^{2} + h^{2})^\frac{3}{2}}$ $\\$ For maximum E, $\frac{dE}{dh} =0 \Rightarrow \frac{l_0 [(R^2 + h^2)^\frac{3}{2} - \frac{3}{2}h \times (R^2 + h^2)^\frac{t}{2} \times 2h ]}{(R^2 + h^2)^3}$ $\\$ $\Rightarrow (R^2 + h^2)^\frac{1}{2}[R^2 + h^2 -3h^2] = 0$ $\\$ $\Rightarrow R^2 - 2h^2 = 0 \Rightarrow h = \frac{R}{\sqrt 2}$

16   Figure $(22-E1)$ shows a small diffused plane source S placed over a horizontal table-top at a distance of $2.4 m$ with its plane parallel to the table-top. The illuminance at the point A directly below the source is $25 lux$. Find the illuminance at a point B of the table at a distance of $1.8 m$ from A.

##### Solution :

17   An electric lamp and a candle produce equal illuminance at a photometer screen when they are placed at $80 cm$ and $20 cm$ from the screen respectively. The lamp is now covered with a thin paper which transmits 49% of the luminous flux. By what distance should the lamp be moved to balance the intensities at the screen again ?