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% .\]