The intensity of light from a source decreases with the distance x from the source, as x 2 1 . In the following picture, the intensity of light from the source S at point A on a curtain is 8.1 U , w

Question

The intensity of light from a source decreases with the distance x from the source, as $\frac{1}{x ^{2}}$ . In the following picture, the intensity of light from the source S at point A on a curtain is 8.1 U , where U is some unit. We then add a big mirror parallel to the curtain at point B . The distances AS and SB are equal, and the points $A, S$ and B are in line, perpendicular to the curtain and the mirror. What is the intensity in U of light reaching point A now?

Options:

Answer:

A

Solution:

$I_{net} = I_{A} + I_{B}$

$I_{B} = \frac{I _{0}}{4 π L ^{2} \times 9}, I_{A} = \frac{P _{0}}{4 π L ^{2}} = 8.1 μ$

\begin{center}

\end{center}

$I_{n e t} = 8.1 μ + \frac{8.1 μ}{9} I_{n e t} = 9 μ$

TestHub · Accepted Answer

I net ​ = I A ​ + I B ​ I B ​ = 4 π L 2 × 9 I 0 ​ ​ , I A ​ = 4 π L 2 P 0 ​ ​ = 8.1 μ \begin{center} \end{center} ​ I n e t ​ = 8.1 μ + 9 8.1 μ ​ ​ I n e t ​ = 9 μ ​

Physics - Geometrical Optics Question with Solution | TestHub

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