A point source, positioned at the origin, emits 628 lumen of...

The luminous intensity $I$ of the point source is given by the total luminous flux divided by the total solid angle:

$I = \dfrac{\Phi}{4\pi} = \dfrac{628}{4 \times 3.14} = 50 \text{ cd}$

The illuminance $E$ on a surface is given by:

$E = \dfrac{I \cos \theta}{r^2}$

where $r$ is the distance from the source and $\theta$ is the angle between the normal to the surface and the direction of the incident light. The light travels along the X-axis, so the angle of incidence is \theta = 37^{}^{{}^{\circ}}.

Substituting the given values:

E = \dfrac{50 \times \cos 37^{}^{{}^{\circ}}}{(1.0)^2}

$E = 50 \times \dfrac{4}{5} = 40 \text{ lux}$