An electric light is placed directly over the centre of a circular plot of lawn 100 m in diameter. Assuming that the intensity of light varies directly as the sine of the angle at which it strikes an illuminated surface and inversely as the square of its distance from its surface, how should the light be hung in order that the intensity may be as great as possible at the circumference of the plot?

Answer

`25sqrt(2)m` above the ground

Explanation

Let C be the center of the circular plot of lawn with a diameter of 100m i.e
CA=CB=50 m

Let o be the positon of light which is direcltly above c also let `angleOAC=theta,(0ltthetaltpi//2)`
Then according to the question the intensity I of the light at the circumcenter of the plot is given by
`I=(ksin theta)/(OA)^(2)=(k sin theta)/(50 sec theta)^(2)`
or `I=[K//(25000)sinthetacos^(2)theta]`
`therefore d(I)/(d theta)=(k//2500)(cos^(3)theta-2 sin^(2)theta cos theta)`
`=(k//2500)cos theta (cos^(2)theta-2 sin^(2) theta)`
for maximum or minimum of `I,dI//d theta=0, tan theta =1 sqrt(2)`
Now we have
`d^(2)I/d theta^(2)=k/2500(-7 sin theta - 2 sin^(2) theta cos theta)`
`=(k//2500) sin theta cos^(2)theta(-7+2 tan^(2)theta)`
When `tan theta =1//sqrt(2),d^(2)I//d theta^(2) is -ve`
Hence ,I (intensity of light) is maximum when tan `theta =1sqrt(2)`
Now we have
`d^(2)I//d theta^(2)=(k//2500)(-7 sin theta cos^(2) theta + sin ^(3) theta)`
when `tan theta =1//sqrt(2),d^(2)I//d theta^(2)`is - ve
Hence I (intensity of light) is maximum when `tan theta =1sqrt(2)` which is the only point of extrema so gives the greatest intensity Thus the required height of the light is
`OC=AC tan theta = 5sqrt(2) =25sqrt(2)m`