
A lamp hanging $4$ metres above the table is lowered by $1$ metre. The illumination on the table:
(A) Decreased by $25\% $
(B) Increased by $25\% $
(C) Decreased by $66.7\% $
(D) Increased by $77.7\% $
Answer
584.1k+ views
Hint:Here we have to find the illumination intensity formula at a point. Then we can find the percentage increase in illumination intensity.
Complete step by step answer:
In photometry, luminous intensity is a standardised model of the sensitivity of the human eye, a measure of the wavelength-weighted force produced by a light source in a particular direction per unit solid angle, based on luminosity function.
Mathematically luminous intensity is given by-
$E = \dfrac{{{I_ \circ }}}{{{d^2}}}$
Given,
A lamp hanging $4$ metres above the table is lowered by $1$ metre.
Illumination intensity at any point is given by $E = \dfrac{{{I_ \circ }}}{{{d^2}}}$
where $d$ is the distance of the point from the source.
When the distance is ${d_1} = 4\,m$
${E_1} = \dfrac{{{I_ \circ }}}{{{4^2}}} = 0.0625{I_ \circ }$
Now when the lamp is lowered by $1\,m$ i.e. distance is ${d_2} = 3\,m$
${E_2} = \dfrac{{{I_ \circ }}}{{{3^2}}} = 0.1111{I_ \circ }$
Percentage increase in intensity of illumination is given by-
$
= \dfrac{{{E_2} - {E_1}}}{{{E_2}}} \times 100 \\
= \dfrac{{0.1111{I_ \circ } - 0.0625{I_ \circ }}}{{0.0625{I_ \circ }}} \times 100 \\
= 77.7\% \\
$
Therefore, the illumination on the table increased by $77.7\% $.
Hence, option D is correct.
Note:While calculating the percentage increase in luminous intensity we have to be careful while putting the values of different intensities otherwise we will get a different answer.And also remember that lux is expressed as the number of lumens falling on a surface. One lux lumen per square metre, linking light to the source’s distance (lux) is the amount of light reflecting from a surface.Depending on the direction, it primarily helps to determine the distribution of the light given off by a lit surface. The term is often used for a point source approximation, i.e. in distances that are wide to the source degree. The S.I unit of luminous intensity is Candela. The luminous flux is the rate at which luminous intensity is measured.
Complete step by step answer:
In photometry, luminous intensity is a standardised model of the sensitivity of the human eye, a measure of the wavelength-weighted force produced by a light source in a particular direction per unit solid angle, based on luminosity function.
Mathematically luminous intensity is given by-
$E = \dfrac{{{I_ \circ }}}{{{d^2}}}$
Given,
A lamp hanging $4$ metres above the table is lowered by $1$ metre.
Illumination intensity at any point is given by $E = \dfrac{{{I_ \circ }}}{{{d^2}}}$
where $d$ is the distance of the point from the source.
When the distance is ${d_1} = 4\,m$
${E_1} = \dfrac{{{I_ \circ }}}{{{4^2}}} = 0.0625{I_ \circ }$
Now when the lamp is lowered by $1\,m$ i.e. distance is ${d_2} = 3\,m$
${E_2} = \dfrac{{{I_ \circ }}}{{{3^2}}} = 0.1111{I_ \circ }$
Percentage increase in intensity of illumination is given by-
$
= \dfrac{{{E_2} - {E_1}}}{{{E_2}}} \times 100 \\
= \dfrac{{0.1111{I_ \circ } - 0.0625{I_ \circ }}}{{0.0625{I_ \circ }}} \times 100 \\
= 77.7\% \\
$
Therefore, the illumination on the table increased by $77.7\% $.
Hence, option D is correct.
Note:While calculating the percentage increase in luminous intensity we have to be careful while putting the values of different intensities otherwise we will get a different answer.And also remember that lux is expressed as the number of lumens falling on a surface. One lux lumen per square metre, linking light to the source’s distance (lux) is the amount of light reflecting from a surface.Depending on the direction, it primarily helps to determine the distribution of the light given off by a lit surface. The term is often used for a point source approximation, i.e. in distances that are wide to the source degree. The S.I unit of luminous intensity is Candela. The luminous flux is the rate at which luminous intensity is measured.
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