A lamp is hanging 1 , m above the center of a | vedclass

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(B) Let the height of the lamp above the center be $h = 1 \, m$. The radius of the table is $r = 0.5 \, m$ (since diameter is $1 \, m$).Illuminance $E

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$(\frac{5}{4})^{3/2}$

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(B) Let the height of the lamp above the center be $h = 1 \, m$. The radius of the table is $r = 0.5 \, m$ (since diameter is $1 \, m$).Illuminance $E$ at a point is given by $E = \frac{I \cos 	heta}{d^2}$,where $I$ is the luminous intensity of the lamp,$	heta$ is the angle of incidence,and $d$ is the distance from the lamp.At the center,$	heta = 0^\circ$,so $\cos 	heta = 1$ and $d = h$. Thus,$E_{center} = \frac{I}{h^2}$.At the edge,the distance $d = \sqrt{h^2 + r^2}$ and $\cos 	heta = \frac{h}{d} = \frac{h}{\sqrt{h^2 + r^2}}$.Thus,$E_{edge} = \frac{I \cdot h}{d^3} = \frac{I \cdot h}{(h^2 + r^2)^{3/2}}$.The ratio is $\frac{E_{center}}{E_{edge}} = \frac{I/h^2}{I \cdot h / (h^2 + r^2)^{3/2}} = \frac{(h^2 + r^2)^{3/2}}{h^3}$.Substituting $h = 1$ and $r = 0.5 = 1/2$:$\frac{E_{center}}{E_{edge}} = \frac{(1^2 + (1/2)^2)^{3/2}}{1^3} = (1 + 1/4)^{3/2} = (5/4)^{3/2}$.

List $I$ (Position of the object)	List $II$ (Magnification)
$(I)$ An object is placed at focus before a convex mirror	$(A)$ Magnification is $-\infty$
$(II)$ An object is placed at centre of curvature before a concave mirror	$(B)$ Magnification is $0.5$
$(III)$ An object is placed at focus before a concave mirror	$(C)$ Magnification is $+1$
$(IV)$ An object is placed at centre of curvature before a convex mirror	$(D)$ Magnification is $-1$
	$(E)$ Magnification is $0.33$

$A$ lamp is hanging $1 \, m$ above the center of a circular table of diameter $1 \, m$. The ratio of illuminances at the center and the edge is:

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