A point source emitting 628 lumen of luminous flux uniformly in all directions is placed at the origin. Calculate the illuminance on a small area placed at (1.0 m, 0, 0) in such a way that the normal to the area makes an angle of `37^@` with the X-axis.

Explanation

To solve the problem, we need to calculate the illuminance on a small area placed at the coordinates (1.0 m, 0, 0) with the normal to the area making an angle of \(37^\circ\) with the X-axis. Here’s a step-by-step solution:

### Step 1: Identify the Given Data
- Luminous flux (\( \Phi \)) = 628 lumens
- Distance from the source (\( r \)) = 1.0 m
- Angle (\( \theta \)) = \( 37^\circ \)

### Step 2: Calculate the Solid Angle
Since the luminous flux is emitted uniformly in all directions, the solid angle (\( \Omega \)) for a point source is given by:

\[ \Omega = 4\pi \text{ steradians} \]

### Step 3: Calculate Luminous Intensity
Luminous intensity (\( I \)) can be calculated using the formula:

\[ I = \frac{\Phi}{\Omega} \]

Substituting the values:

\[ I = \frac{628 \text{ lumens}}{4\pi} \approx \frac{628}{12.5664} \approx 50 \text{ candela} \]

### Step 4: Calculate the Illuminance
The illuminance (\( E \)) at a point is given by the formula:

\[ E = \frac{I \cdot \cos \theta}{r^2} \]

Substituting the values we have:

\[ E = \frac{50 \cdot \cos(37^\circ)}{1^2} \]

### Step 5: Calculate \( \cos(37^\circ) \)
From trigonometric tables or a calculator, we find:

\[ \cos(37^\circ) = \frac{4}{5} = 0.8 \]

### Step 6: Substitute and Calculate Illuminance
Now substituting \( \cos(37^\circ) \) into the illuminance formula:

\[ E = 50 \cdot 0.8 = 40 \text{ lux} \]

### Final Answer
The illuminance on the small area is **40 lux**.

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