Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

A Point Source Emitting 628 Lumen of Luminous Flux Uniformly in All Directions is Placed at the Origin. - Physics

Sum

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.

Solution

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.

Concept: Light Process and Photometry
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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 22 Photometry
Exercise | Q 9 | Page 455
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