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 ____________ .
Options
0.5%
1%
2%
4%
Solution
2%
Illuminance is given by
\[E = \frac{I_o \cos\theta}{r^2}\]
\[\theta = 0^0 \]
\[\frac{\Delta r}{r} = 1 \% \]
\[E = \frac{I_o}{r^2}\]
Differentiating,
\[dE = - 2\frac{I_o}{r^3}dr\]
As approximation differentials are replaced by \[\Delta, \]
\[\Delta E = - 2\frac{I_o}{r^3}\Delta r\]
\[ \Rightarrow \Delta E = - 2\frac{I_o}{r^2}\left( \frac{\Delta r}{r} \right)\]
\[ \Rightarrow \Delta E = - 2E\left( \frac{\Delta r}{r} \right)\]
\[ \Rightarrow \frac{\Delta E}{E} = - 2\left( \frac{\Delta r}{r} \right)\]
\[ \Rightarrow \frac{\Delta E}{E} = - 2 \times 1 \% = - 2 \% \]
Since, negative sign implies decrease; hence, illuminance decreases by 2% .\]
