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(B) Let the height of the light source be $h$ and the radius of the table be $R$. The intensity of light $I$ at the edge of the table is given by the

Vedclass · Accepted Answer

$R/\sqrt{2}$

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(B) Let the height of the light source be $h$ and the radius of the table be $R$. The intensity of light $I$ at the edge of the table is given by the formula $I = \frac{L \cos 	heta}{d^2}$,where $L$ is the luminous intensity of the source,$d$ is the distance from the source to the edge,and $	heta$ is the angle between the normal and the light ray.Here,$d^2 = h^2 + R^2$ and $\cos 	heta = \frac{h}{d} = \frac{h}{\sqrt{h^2 + R^2}}$.Substituting these,we get $I = \frac{Lh}{(h^2 + R^2)^{3/2}}$.To find the height $h$ for maximum intensity,we differentiate $I$ with respect to $h$ and set it to zero: $\frac{dI}{dh} = L \left[ \frac{(h^2 + R^2)^{3/2} - h \cdot \frac{3}{2}(h^2 + R^2)^{1/2} \cdot 2h}{(h^2 + R^2)^3} ight] = 0$.This simplifies to $(h^2 + R^2) - 3h^2 = 0$,which gives $R^2 = 2h^2$.Therefore,$h = R/\sqrt{2}$.

$A$ small source of light is to be suspended directly above the centre of a circular table of radius $R$. What should be the height of the light source above the table so that the intensity of light is maximum at the edges of the table compared to any other height of the source?

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