A bulb of 100 text{ W} is hanging at a height | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) 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

$5\sqrt{5} I_0$

Vedclass · Answer

(D) 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 between the normal to the surface and the light ray.At the centre $B$,the distance $r_B = 1 	ext{ m}$ and the angle $	heta_B = 0^\circ$. Thus,$\cos 	heta_B = 1$.$I_B = \frac{L}{1^2} = L$ ... $(i)$At the rim $C$,the distance $r_C = \sqrt{1^2 + 2^2} = \sqrt{5} 	ext{ m}$. The angle $	heta_C$ is such that $\cos 	heta_C = \frac{1}{\sqrt{5}}$.$I_C = I_0 = \frac{L \cos 	heta_C}{r_C^2} = \frac{L 	imes (1/\sqrt{5})}{(\sqrt{5})^2} = \frac{L}{5\sqrt{5}}$ ... $(ii)$From $(i)$ and $(ii)$,$L = 5\sqrt{5} I_0$.Therefore,the intensity at the centre is $I_B = 5\sqrt{5} I_0$.

$A$ bulb of $100 \text{ W}$ is hanging at a height of $1 \text{ m}$ above the centre of a circular table of diameter $4 \text{ m}$. If the intensity at a point on its rim is $I_0$,then the intensity at the centre of the table will be:

Similar Questions

If the luminous efficiency of a lamp is $2 \text{ lumen/watt}$ and its luminous intensity is $42 \text{ candela}$,then the power of the lamp is.......$W$

If $\hat{n}_1$ is the unit vector along the incident ray,$\hat{n}_2$ is the unit vector along the normal,and $\hat{n}_3$ is the unit vector along the reflected ray,then which of the following must be true?

Which of the following phenomena are involved in the formation of a rainbow?

Shown in the figure is a convergent lens placed inside a cell filled with a liquid. The lens has a focal length of $+20\,cm$ when in air and its material has a refractive index of $1.50$. If the liquid has a refractive index of $1.60$,the focal length of the system is......$cm$.

The camera lens has an aperture of $f$ and the exposure time is $(1/60) \, s$. What will be the new exposure time if the aperture becomes $1.4 \, f$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module