Light originating from a point source strikes a screen. If the...

The illuminance $E$ produced by a point source at a distance $r$ is given by $E = \dfrac{I}{r^2}$, where $I$ is the luminous intensity of the source.

Taking the natural logarithm on both sides, we get $\ln E = \ln I - 2 \ln r$.

Differentiating for small changes, we obtain $\dfrac{\Delta E}{E} = -2 \dfrac{\Delta r}{r}$.

Given that the separation $r$ is increased by $1\%$, we have $\dfrac{\Delta r}{r} \times 100 = 1\%$.

Therefore, the percentage change in illuminance is $\dfrac{\Delta E}{E} \times 100 = -2 \left( \dfrac{\Delta r}{r} \times 100 \right) = -2(1\%) = -2\%$.

The negative sign indicates that the illuminance decreases. Thus, the illuminance decreases by approximately $2\%$.