Question

An electric lamp is fixed at the ceiling of a circular tunnel as shown in figure. What is the ratio of the intensities of light at base A and at a point B on the wall

• A. 1:2
• B. $2:\sqrt{3}$
• C. $\sqrt{3:1}$
• D. 1:$\sqrt{2}$

Answer 1

Answer: (D)

1:$\sqrt{2}$

Answer 2

Let R be the radius of the tunnel. Then SA =2R and $SB = \sqrt{2}R$, intensity of illumination at A is

$E_{A} = \frac{I}{(2R)^{2}} = \frac{I}{4R^{2}}$

As shown above, the intensity of illumination at B will be

$E_{B} = \frac{I}{(\sqrt{2}R)^{2}}\cos\theta = \frac{I\cos 45º}{2R^{2}} = \frac{I}{2\sqrt{2}R^{2}}$

Where $\theta$ is the angle $\angle BSO$

Therefore $\frac{E_{A}}{E_{B}} = \frac{2\sqrt{2}}{4} = \frac{1}{\sqrt{2}}$