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प्रश्न
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.
उत्तर
Given,
Luminous flux = 628 lumen
Angle made by the normal with the x axis (θ) = 37°
Distance of point, r = 1 m
Since the radiant flux is distributed uniformly in all directions, the solid angle will be 4π.
∴ Luminous intensity, l = `"Luminous flux"/"Solid angle"`
`=628/(4pi)=50 "candela"`
I lluminance (E) is given by,
`E=l cos (theta/r^2)`
On substituting the respective values we get,
`E=50xxcos37^o/1^2`
`50xx"(4/5)"/1=40 "lux"`
So, the illuminance on the area is 40 lux.
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