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Question

In the figure shown, O is a point source of light and S is a screen placed at a distance L from the source. The intensity of light at point A on the screen due to the source is 81I , where I is some unit. Now a large mirror (M) is placed behind the source at a distance L from it. The mirror reflects 100% of the light energy incident on it. The intensity at point A is nI. The value of n is (integer only)


A
90
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B
90.00
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C
90.0
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Solution

Imagine the image of O to be another point source. The mirror reflects the light as if it is diverging from the image. The wavefronts around a point source are spherical in shape and the intensity falls off with distance (x) from the source as I1x2Image as a source, is at a distance 3L from the point A.

Intensity due to it will be 19th of that caused by the original source.

Intensity at A=81I+81I9=90I

n=90

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