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(A) Let $L$ be the luminous intensity of the bulb.The illuminance $I$ at a point is given by $I = \frac{L \cos 	heta}{r^2}$,where $r$ is the distance

Vedclass · Accepted Answer

$(17/13)^{3/2}$

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(A) Let $L$ be the luminous intensity of the bulb.The illuminance $I$ at a point is given by $I = \frac{L \cos 	heta}{r^2}$,where $r$ is the distance from the source and $	heta$ is the angle of incidence.For point $A$ (center of the edge):The horizontal distance from the center is $2 	ext{ m}$. The vertical height is $3 	ext{ m}$.The distance $r_A = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} 	ext{ m}$.$\cos 	heta_A = \frac{3}{\sqrt{13}}$.$I_A = \frac{L \cos 	heta_A}{r_A^2} = \frac{L 	imes (3/\sqrt{13})}{13} = \frac{3L}{13^{3/2}}$.For point $B$ (corner of the table):The horizontal distance from the center is $\sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} 	ext{ m}$. The vertical height is $3 	ext{ m}$.The distance $r_B = \sqrt{3^2 + (2\sqrt{2})^2} = \sqrt{9 + 8} = \sqrt{17} 	ext{ m}$.$\cos 	heta_B = \frac{3}{\sqrt{17}}$.$I_B = \frac{L \cos 	heta_B}{r_B^2} = \frac{L 	imes (3/\sqrt{17})}{17} = \frac{3L}{17^{3/2}}$.The ratio of intensities is $\frac{I_A}{I_B} = \frac{3L / 13^{3/2}}{3L / 17^{3/2}} = \left(\frac{17}{13}ight)^{3/2}$.

$A$ $60 \text{ W}$ bulb is hung over the center of a table of size $4 \text{ m} \times 4 \text{ m}$ at a height of $3 \text{ m}$. The ratio of the intensities of illumination at a point on the center of the edge $(A)$ and at the corner of the table $(B)$ is:

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