Tag: physics

Questions Related to physics

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

 The luminous flux emitted by the sun will be

luminous intensity measurements physics

  1. $4.4\times10^{25}$ lm

  2. $4.43\times10^{26}$ lm

  3. $4.43\times10^{27}$ lm

  4. $4.43\times10^{28}$ lm

Show answer
Correct Option: D
Explanation:

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

$ I = 3.53\times 10^{27}$
$ Solid angle  = 4 \pi$

So Luminous flux = $3.53\times10^{27}\times4 \pi$
= $ 4.43\times10^{28} lm$

Answer. D) $ 4.43\times10^{28} lm$

A screen recieves $3$ watt of radiant flux of wavelength $6000$ $\mathring{A}$. One lumen is equivalent to $1.5\times10^{-3}$ watt of monochromatic light of wavelength $5500$ $\mathring{A}$ is $1.00$, then the luminous flux of the source is ___

luminous intensity measurements physics

  1. $4\times10^{3}$ $lm$

  2. $3\times10^{3}$ $lm$

  3. $2\times10^{3}$ $lm$

  4. $1.37\times10^{3}$ $lm$

Show answer
Correct Option: D
Explanation:

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

relative luminosity at 6000 angstorm is 0.685   
relative luminosity at 5500 angstorm is 1

luminous efficiency  = $\dfrac{0.685}{1}$  

therefore luminous flux = luminous efficiency $\times$ radiant flux = 0.685 $\times$2000 = $1.37 \times 10^3lm$

The luminous intensity of a $100$ W unidirectional bulb is $100$ candela. The total luminous flux emitted from the bulb will be

luminous intensity measurements physics

  1. $100\pi$

  2. $200\pi$

  3. $300\pi$

  4. $400\pi$

Show answer
Correct Option: D
Explanation:

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

A point source of $100$candela is held $5$$m$ above a sheet of blotting paper which reflects $75$ % of light incident upon it. The illuminance of blotting paper is 

luminous intensity measurements physics

  1. $4$ phot

  2. $4$ lux

  3. $3$ phot

  4. $3$ lux

Show answer
Correct Option: B
Explanation:
Illuminance = $\dfrac{L}{d^{2}}$
$\dfrac{100}{5^{2}}$ = 4 lux

Answer. B) 4 lux

Inverse square law for illuminance is valid for 

luminous intensity measurements physics

  1. isotropic point source

  2. cylindrical source

  3. search light

  4. all types of sources

Show answer
Correct Option: D
Explanation:

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