An electric bulb is suspended over a table at a height of 1 m . The...

The illuminance directly beneath the bulb is given by $E_0 = \dfrac{I}{h^2}$

Given $E_0 = 40 \text{ lux}$ and $h = 1 \text{ m}$, $I = 40 \text{ cd}$.

For a point on the table at a distance $x = 1 \text{ m}$ from the point directly beneath the bulb, the distance from the bulb is $r = \sqrt{h^2 + x^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \text{ m}$.

The angle of incidence $\theta$ satisfies $\cos \theta = \dfrac{h}{r} = \dfrac{1}{\sqrt{2}}$.

The illuminance at this point is $E = \dfrac{I \cos \theta}{r^2}$

Substituting the values, $E = \dfrac{40 \times \left(\dfrac{1}{\sqrt{2}}\right)}{(\sqrt{2})^2} = \dfrac{40}{2\sqrt{2}} = 10\sqrt{2}$

Since $\sqrt{2} \approx 1.414$, $E \approx 10 \times 1.414 = 14.14 \text{ lux}$.

The illuminance is approximately $14 \text{ lux}$.