Tag: luminous intensity

Questions Related to luminous intensity

Light from a point source falls on a screen. If the separation between the source and the screen is increased by 1% the illuminance will decrease 

luminous intensity measurements physics

  1. $0.5$ %

  2. $1$ %

  3. $2$ %

  4. $4$ %

Show answer
Correct Option: C
Explanation:

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

$\displaystyle \frac{\triangle E}{E}$ $=\displaystyle\ \frac{2\triangle r}{r}$ 

$\Rightarrow$ $2$ ($1$%) $= 2$%

$1$ % of light of a source with luminous intensity $50 $candela is incident on a circular surface of radius $10 cm$. The average illuminance of the surface is 

luminous intensity measurements physics

  1. $100$ lux

  2. $200$ lux

  3. $300$ lux

  4. $400$ lux

Show answer
Correct Option: B
Explanation:

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

                    =$ \dfrac {50 \times 4 \pi}{100\times \pi\times0.1\times0.1} = 200 lux$

Answer B) $200 \ lux$

The illumination produced by A is balanced by B on the screen when B is 60 cm apart from the screen. A smoked glass plate is placed in front of A and to balance the illumination B is to move 15cm further away. Find the transmission coefficient of the smoked glass

luminous intensity measurements physics

  1. $0.36$

  2. $0.64$

  3. $0.49$

  4. $0.51$

Show answer
Correct Option: B
Explanation:

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

Two light sources of $8 Cd$ and $12 Cd$ are placed on the same side of the photometer screen at a distance of $40 cm$ from it. Where should a $80 Cd$ source be placed to balance the illuminance?

luminous intensity measurements physics

  1. $40 cm$

  2. $60 cm$

  3. $20 cm$

  4. $80 cm$

Show answer
Correct Option: D
Explanation:

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


Therefore
$I _1+I _2=I$

$\dfrac {8} {0.4^{2}} + \dfrac{12} {0.4^{2}} = \dfrac{80}{d^{2}}$

=>$\dfrac {20} {0.16} = \dfrac{80}{d^{2}}$

=> d = 0.8 m =80 cm

Answer. D) 80 cm

The luminous intensity of a light source is $300 Cd$. The illuminance of a surface lying at a distance of $10$ $m$ from it will be if light falls normally on it 

luminous intensity measurements physics

  1. $30$ lux

  2. $3$ lux

  3. $0.3$ lux

  4. $0.03$ lux

Show answer
Correct Option: B
Explanation:

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 $R = 10 m$
$\implies E = \dfrac {300}{10^2} = 3   lux$

The luminous intensity of a light source is $500 Cd$. The illuminance of a surface distant $10m$ from it, will be if light falls normally on it

luminous intensity measurements physics

  1. $5$ lux

  2. $10$ lux

  3. $20$ lux

  4. $40$ lux

Show answer
Correct Option: A
Explanation:

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 $R = 10 m$
$\implies E = \dfrac {500}{10^2} = 5    lux$

A lamp is hanging at a height of $4$ $m$ above a table. The lamp is lowered by $1$ $m$. The percentage increase in illuminance is 

luminous intensity measurements physics

  1. $40$ %

  2. $64$ %

  3. $78$ %

  4. $92$ %

Show answer
Correct Option: C
Explanation:

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


The only change when lamp moves from 4 m to 3 m is its d
Illuminance at 4 m = Power / 16
Illuminance at 3 m = Power / 9

Increase in Illuminance = $\dfrac{power}{9} - \dfrac{power}{16}$ = $ \dfrac{7 Power}{144}$

So % increase = $ \dfrac{7 Power}{144} * \dfrac{16 * 100}{Power}$ = 77.77 % 

Answer. C) 78 %

A photoprint is required to be placed in front of $100$ $Cd$ lamp at a distance of $0.5$ $m$ for $25$ sec for good impression. If it is to be placed in front of a $400$ $Cd$ lamp for $36$ sec for the same impression then the distance of the print from the lamp will be 

luminous intensity measurements physics

  1. $0.5$ $m$

  2. $1.0$ $m$

  3. $1.2$ $m$

  4. $1.5$ $m$

Show answer
Correct Option: C
Explanation:

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


Let the required distance be d m
So,
$\dfrac{100 * 25}{0.5^{2}} = \dfrac{400 * 36}{d^{2}}$
=> d = 1.2 m

Answer. C) 1.2 m

Two lamps of luminous intensity of $8$ $Cd$ and $32$ $Cd$ respectively are lying at a distance of $1.2$ $m$ from each other. Where should a screen be placed between two lamps such that its two faces are equally illuminated due to the two sources?

luminous intensity measurements physics

  1. $10$ $cm$ from $8$ $Cd$ lamp

  2. $10$ $cm$ from $32$ $Cd$ lamp

  3. $40$ $cm$ from $8$ $Cd$ lamp

  4. $40$ $cm$ from $32$ $Cd$ lamp

Show answer
Correct Option: C
Explanation:

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

Distance between 32 Cd lamp and screen $=$ $ (120-x) m$

Under equal illumination,
$ \dfrac{8}{x^{2}} = \dfrac{32}{(120-x)^{2}}$
=> x $=$ 40 cm

So the distance from 8 Cd lamp is 40 centimeters.

At what distance should a book be placed from a $50 Cd$ bulb so that the illuminance on the book becomes $2lm $ $m^{-2}$

luminous intensity measurements physics

  1. $1$m

  2. $5$m

  3. $10$m

  4. $50$m

Show answer
Correct Option: B
Explanation:

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 $E = 2 lm  m^{-2}$
$\implies R = \sqrt {\dfrac {50}{2}} = 5   m$