- Download PDF
"A light is hung from rafter at a height h m above the floor , providing a circle of illumination. the illuminance E (in lumens/ squares meter or lm/`m^(2)` ) at ant point p on the floor varies directly as the cosine of angle `theta` at which the rays leave the light source toward P, and inversely as the square of the height h of the light source, using k as the constant of variation, write expression for illuminance in term k, h, and `theta`. If the distance from the light source to point P is 13 m
write the expression in illuminance in terms of k and h alone.
for k = 15,600, what is the height of the light source, if E = 100lm/m^2?
1 Answer | Add Yours
Given, Illuminance E, at any point P varies directly as the cosine of angle `theta` at which the rays leave the light source toward P.
Mathematically, `E prop costheta`
Again, illuminance E, at any point P varies inversely as the square of the height h of the light source from the point P (refer to the attached diagram).
`E prop 1/h^2`
Combining these two relations,
`E prop costheta/h^2`
`rArr ` `E=(k*costheta)/h^2` --- (i) where, k is the constant of proportionality
This is the expression for illuminance in terms of k, h and theta.
Now, triangle PCL is a right triangle, `costheta=h/(PL)=h/13`
Therefore, `E= k*costheta`
`rArr E=(k* h)/13*1/h^2=k/(13h)` --- (ii)
This is the expression for illuminance in terms of k and h.
Given k=15600, E=100 lm/m^2, plugging in the values in equation (ii),
`rArr` `h=12 m`
Therefore, for k = 15,600 and E =100 lm/m^2, the height of the light source, is 12 m.
We’ve answered 319,856 questions. We can answer yours, too.Ask a question