Tag: measurements

Questions Related to measurements

The luminous efficiency of the bulb in lumen/watt, if luminous intensity of a $100$ watt unidirectional bulb is $100$ candela, is

luminous intensity measurements physics

  1. $12$

  2. $12.56$

  3. $13$

  4. $15$

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Correct Option: B
Explanation:

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

An electric bulb is suspended at a vertical height $2\ m$ from the centre of a square table of side $2\ m$. If the luminous intensity of bulb is $60\ cd$ (candela), then the illumination at one corner of the table is:

luminous intensity measurements physics

  1. $8.16\ cd/m^{2}$

  2. $6.24\ cd/m^{2}$

  3. $9.25\ cd/m^{2}$

  4. $8.72\ cd/m^{2}$

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Correct Option: A

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

luminous intensity measurements physics

  1. $100 \pi$ lumen

  2. $200 \pi$ lumen

  3. $300 \pi$ lumen

  4. $400 \pi$ lumen

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Correct Option: D
Explanation:

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


Answer. D) $400\pi$ lumen

The intensity produced by a long cylindrical light source at a small distance $r$ from the source is proportional to

luminous intensity measurements physics

  1. $\displaystyle \dfrac{1}{r^2}$

  2. $\displaystyle \dfrac{1}{r^3}$

  3. $\displaystyle \dfrac{1}{r}$

  4. None of these

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Correct Option: C
Explanation:

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


$ I=\dfrac { P }{ S } =\dfrac { P }{ 2\pi rl } \Rightarrow I\propto \dfrac { 1 }{ r } $

Two light sources with equal luminous intensity are lying at a distance of 1.2 m from each other. Where should a screen be placed between them such that illuminance on one of its faces is four times that on another face?

luminous intensity measurements physics

  1. 0.2 m

  2. 0.4 m

  3. 0.8 m

  4. 1.6 m

Show answer
Correct Option: C
Explanation:

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

The symbol of candela is __________.

luminous intensity measurements physics

  1. Cd

  2. C

  3. Can

  4. Ca

Show answer
Correct Option: A
Explanation:
The symbol of candela is Cd.

Light from a source is analysed by an analyser. When the analyser is rotated, the intensity of the emergent light.

luminous intensity measurements physics

  1. Does not vary

  2. Remains uniformly dark

  3. Varies between maximum and zero

  4. Varies between maximum and minimum

Show answer
Correct Option: A
Explanation:

Let the intensity of the unpolarised light be $I _o$.
An analyser is used to polarized the light in one direction.
Intensity of light after passing through the analyser  $I' = \dfrac{I _o}{2}$
Intensity of light after passing through analyser remains constant $I _o/2$ even if the analyser is rotated by some angle.

 If light falls on the surface at an angle of $60^{0}$, then illuminance will be____

luminous intensity measurements physics

  1. $12$ lux

  2. $6$ lux

  3. $3$ lux

  4. $1.5$ lux

Show answer
Correct Option: D
Explanation:

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 $I = 300 cd$, $R = 10 m$, $\alpha = 60^o$
$\implies E = \frac {300}{10^2}    cos  60^o= 1.5   lux$

A lamp is hanging along the axis of a circular table of radius r. At what height should the lamp be placed above the table, so that the illuminance at the edge of the table is $\displaystyle \frac{1}{8}$ of that at its centre?

luminous intensity measurements physics

  1. r/2

  2. r/$\sqrt{2}$

  3. r/3

  4. r/$\sqrt{3}$

Show answer
Correct Option: D
Explanation:

$E _2 = \displaystyle \frac{1}{8} E _1$ 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 $r^2 + h^2 = 4h^2$
$h = r / \sqrt{3}$