Question

A lamp is hanging at a height of 4m above a table. The lamp is lowered by 1m. The percentage increase in illuminance is
a. 40%
b. 64%
c. 78%
d. 92%

Answer 1

Hint: To solve such types of problems one must keep in mind the nature of the relationship, the light intensity or illuminance and the distance share with each other. The relation between the illuminance and distance is of inverse square type,
$E \propto \dfrac{1}{{{d^2}}}$
where, E is illuminance and d is the distance. The above relationship between the light intensity and distance explains why the intensity of the light gets diminished when a source of light moves away. You may have experienced this phenomenon while travelling on road at night as a car approaches you the light intensity increases.

Complete step by step answer:
We know that the illuminance of a lamp at a distance of d metres can be calculated as,
${\text{Illuminance}} = {\text{Illuminating}}{\text{ }}{{power}} \times \dfrac{{\operatorname{Cos} \theta }}{{{d^2}}}$
The only change when lamp moves from 4 m to 3 m is its d

Now, Illuminance of lamp at 4 m can be obtained as follows,
${{{E}}_{{{4}}}} = {\text{ }}\dfrac{{{{Power}}{\text{ }}}}{{{{{4}}^{{2}}}}} = \dfrac{{{{Power}}{\text{ }}}}{{{{16}}}}$

Also, Illuminance of lamp at 3 m can be obtained as follows,
${{{E}}_{3}} = {\text{ }}\dfrac{{{{Power}}{\text{ }}}}{{{3^{{2}}}}} = \dfrac{{{{Power}}{\text{ }}}}{9}{\text{ }}$
∴ Increase in Illuminance $({{\Delta E}})$ is the difference between the illuminance of lamp at 3m and the illuminance of the lamp at 4m, which is given below,
${{\Delta E}} = {{{E}}_{{{3}}\;}} - {\text{ }}{{{E}}_{{{4}}}}$
${{\Delta E}}{\text{ = }}\dfrac{{{{Power}}{\text{ }}}}{9} - \dfrac{{{{Power}}{\text{ }}}}{{16}}$
${{\Delta E}}{\text{ = }}\dfrac{{7{\text{ }}}}{{144}}{{Power}}$

Now, the percentage increase in the illuminance of the lamp can be calculated using illuminance of lamp at 4m as the reference value and it is given by the expression,
 $\% {\text{Increase in Illuminance}} = \dfrac{\text{Increase in illuminance}}{\text{Illuminance at 4m}} \times {{100}}$
i.e. $\% \Delta {{E}}{\text{ }} = \dfrac{{{{\Delta E}}}}{{{{{E}}_{{{4}}}}}} \times {{100}}$

$\% {{\Delta E}}{\text{ }} = \dfrac{{{{7}} \times {{Power}}}}{{{{144}}}} \times \dfrac{{{{16}}{\text{ }} \times {\text{ }}{{100}}}}{{{{Power}}}}$
$\% {{\Delta E}}{\text{ }} = {{77}}.{{77}}\% \approx {{78}}\% $
The percentage increase in the illuminance of lamp due to its lowering is approximately equal to 78%.

Hence, the correct answer is option (C).

Note: While solving such types of questions students may confuse the inverse square relation with inverse relation which will drastically change the answer and most probably the question creator will make sure to include an option that will be in agreement with this incorrect answer. So always remember the following equation, $E \propto \dfrac{1}{{{d^2}}}$. So that such type of confusion can be avoided in the examination.