# Calculate where on the floor underneath the fixtures the intensity of the direct light will be most intense. See below for more explanation.Light shines from the fixture through an arc of angle (θ...

Calculate where on the floor underneath the fixtures the intensity of the direct light will be most intense. See below for more explanation.

Light shines from the fixture through an arc of angle (θ M) measured up from the vertical. The bulbs that are installed in the fixture are designed so that the intensity of the light varies as the sine of the angle θ from the vertical changes (θ reachingzero at either end of the light emitting aperture) In addition, the intensity of the light hitting the floor varies inversely as the square of the distance from the fixture to that point change. Therefore, the intensity of the light reaching P is the product of these two effects.

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It is given that the angle `theta` made by a ray of light emitted from the fixture reaches a value equal to 0 at either end of the light emitting aperture. This is possible when `theta` is the magnitude of the angle made by the ray with the horizontal in either direction. The intensity of light varies directly with the sine of the angle `theta` . Also, the intensity varies inversely with the square of the distance of the point from the fixture.

If the fixture is at a vertical distance h from the floor, it can be considered to be the center of a circle with radius h. It is seen that the point on the floor that lies closest to the fixture is directly below it as all other points lie outside the circle. Also, the angle `theta` made by a ray of light falling on this point is equal to 90 degrees, and sin 90 = 1, the maximum value that the sine function can take.

**This gives the point on the floor underneath the fixture where the intensity of light is most intense as the point lying directly below the fixture.**