Question

A screen receives 3watt of radiant flux of wavelength $6000A{}^\circ $. One lumen is equivalent to $1.5\times {{10}^{-3}}watt$ of monochromatic light of wavelength $5550A{}^\circ $. If relative luminosity for $6000A{}^\circ $ is 0.685 while that for $5550A{}^\circ $is 1.00, then the luminous flux of the source is _____
A. $4\times {{10}^{3}}lm$
B. $3\times {{10}^{3}}lm$
C. $2\times {{10}^{3}}lm$
D.$1.37\times {{10}^{3}}lm$

Answer 1

Hint: We are given the value of one lumen equivalent power, using that convert the given radiant flux of source into lumen. Now you could find the luminous efficiency from the given values of luminous intensities. Recall that the luminous flux is given by the product of luminous efficiency and radiant flux and thus find the answer.

Formula used:
Luminous flux = luminous efficiency × radiant flux

Complete answer:
We are told in the question that a screen is receiving 3watt of radiant flux having wavelength. Now let us convert this value of radiant flux into lumens. We are also given that that 1lumen is equivalent to $1.5\times {{10}^{-3}}watt$ of monochromatic light of wavelength$5550A{}^\circ $.
$1.5\times {{10}^{-3}}watt\to 1lumen$
$\Rightarrow 3watt\to \dfrac{3}{1.5\times {{10}^{-3}}}$
Therefore, radiant flux in lumen is given by,
$\dfrac{3}{1.5\times {{10}^{-3}}}=2\times {{10}^{3}}lm$ ………………………………… (1)
We are also given the relative luminosity for both wavelengths $5550A{}^\circ $ and $6000A{}^\circ $ as 1 and 0.685 respectively.
The relative luminosity of a particular wavelength of light is the ratio of luminous flux of that wavelength to the luminous flux of $5550A{}^\circ $wavelength at same power.
We know that, luminous efficiency is given by the ratio of total luminous flux radiated by any source to the total radiant flux from that source, that is,
Luminous flux = luminous efficiency × radiant flux …………………………………… (2)
But luminous efficiency is also defined as average spectral sensitivity of human perception of brightness and could be given by the ratio of the given relative velocities of wavelengths.
Luminous efficiency = $\dfrac{0.685}{1}$ …………………………….. (3)
Substituting (1) and (3) in (2), we get,
Luminous flux = $\dfrac{0.685}{1}\times 2\times {{10}^{3}}=1.37\times {{10}^{3}}lm$
Therefore, we get the luminous flux of the source as $1.37\times {{10}^{3}}lm$

Hence the answer to the question is option D.

Note:
Luminous flux could be defined as the measure of total amount of visible light emitted by a lamp. It is very different from radiant flux as it is the measurement of all electromagnetic radiation emitted, that is, the total amount of objective light. While, luminous flux is the amount of light that is sensed by human eyes.