Tag: measurements

Questions Related to measurements

A lamp of $250$ candle power is hanging at a distance of $6$m from a wall. The illuminace at a point on the wall at a minimum distance from the lamp will be 

luminous intensity measurements physics

  1. $9.64$ lux

  2. $4.69$ lux

  3. $6.94$ lux

  4. none of these

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Correct Option: D
Explanation:

1 Candle power = 0.981 Candela

Luminous intensity = 250*0.981 Candela

Minimum distance of lamp from wall = 6 m

Illuminance = $L/d^{2}$
= $250*0.981/6^{2}$
= 6.8125 lux

Answer. D) none of these

Light from a lamp is falling normally on a surface distant $10$ m from the lamp and the luminous intensity on it is $10$lux. In order to increase the intensity $9$ times, the surface will have to be placed at a distance of 

luminous intensity measurements physics

  1. $10$ $m$

  2. $\displaystyle\ \frac{10}{3}$ $m$

  3. $\displaystyle\ \frac{10}{9}$ $m$

  4. $10\times9$ $m$

Show answer
Correct Option: B
Explanation:

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

Since light falls normally $Cos \theta = 1$
$10 = \dfrac {L} {100}$ => L = 1000Cd

$90 = \dfrac {L} {d^{2}}$ 
 $\longrightarrow d^{2} = 1000/90$
$d= \dfrac{10}{3}$

Answer B) $\dfrac{10}{3} m$

An electric bulb of luminous intensity I is suspended at a height h from the center of the table having a circular surface diameter $2r$, the illuminace at the center of the circular disc will be

luminous intensity measurements physics

  1. $\displaystyle\ \frac{I}{r^{2}}$

  2. $\displaystyle\ \frac{I}{r}$

  3. $\displaystyle\ \frac{I}{h^{2}}$

  4. $\displaystyle\ \frac{I}{h}$

Show answer
Correct Option: C
Explanation:

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

The illuminance of a surface distance $10$ m from a light source is $10$ lux. The luminous intensity of the source for normal incidence will be 

luminous intensity measurements physics

  1. $10^{1} Cd$

  2. $10^{2} Cd$

  3. $10^{3} Cd$

  4. none of these

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Correct Option: C
Explanation:

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

If the distance of surface from light source is doubled then the illuminance will become

luminous intensity measurements physics

  1. $\displaystyle\ \frac{1}{2}$ times

  2. $2$ times

  3. $\displaystyle\ \frac{1}{4}$ times

  4. $4$ times

Show answer
Correct Option: C
Explanation:

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

The one parameter that determines the brightness of a light source sensed by an eye is

luminous intensity measurements physics

  1. energy of light entering the eye per second

  2. wavelength of the light

  3. total radiant flux entering the eye

  4. total luminous flux entering the eye

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Correct Option: D
Explanation:
Luminous flux measures the perceived power of light by adjusting the varying sensitivity to various frequencies of the human eye.

Radiant flux measures the total power of the electromagnetic radiation and does not account for eye sensitivity.

Similarly energy of light entering does not account for the fact that wavelengths beyond visible range does not contribute to brightness.

Wavelength of light alone does not give any idea about the brightness of light either.

Answer. D) total luminous flux entering the eye.

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 (nearly) by

luminous intensity measurements physics

  1. $0.5\%$

  2. $1 \%$

  3. $2\%$

  4. $4\%$

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Correct Option: C
Explanation:

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

A battery-operated torch is adjusted to send an almost parallel beam of light. It produce an illuminance of $40 \ lux$ when light falls on a wall $2 m$ away. The illuminance produced when it falls on a wall $4 m$ away is close to

luminous intensity measurements physics

  1. $40\ lux$

  2. $20\ lux$

  3. $10\ lux$

  4. $5\ lux$

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Correct Option: A
Explanation:

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


Since the beam of light is parallel, so the area does not change with distance. Also the power of torch being same, the luminous power remains same.

Therefore illuminance remains same on a wall 2 m away and 4 m away.

Illuminance produced on a wall 4 m away = 40 lux

Answer. A) 40 lux

The brightness producing capacity of a source

luminous intensity measurements physics

  1. does not depend on its power

  2. does not depend on the wavelength emitted

  3. depends on its power

  4. depends on the wavelength emitted.

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Correct Option: C,D
Explanation:

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

A lamp of luminous intensity $20$ $Cd$ is hanging at a height of $40$ $cm$ from the center of a square table of side $60$ $cm$. The illuminance at the centre of the table will be 

luminous intensity measurements physics

  1. $100$ lux

  2. $125$ lux

  3. $150$ lux

  4. none of these

Show answer
Correct Option: B
Explanation:

Luminous intensity, L = 20 Cd

Distance of source from center, d = 40 cm
Light from lamp hits the center normal ($\theta = 0 ^ {\circ}$)

Lambert's law, $I = L  Cos \theta / d^{2}$
$\longrightarrow \ I = 20  Cos 0 ^{\circ}/ (0.4)^{2}$
$\longrightarrow I = 125 lux$

Answer. B) 125 lux