In the Figure, an isotropic point source of light S is positioned at distance d
ID: 1285282 • Letter: I
Question
In the Figure, an isotropic point source of light S is positioned at distance d from a viewing screen A and the light intensity IP at point P (level with S) is measured. Then a plane mirror M is placed behind S at distance d. By how much is IP multiplied by the presence of the mirror
In the Figure, an isotropic point source of light S is positioned at distance d from a viewing screen A and the light intensity IP at point P (level with S) is measured. Then a plane mirror M is placed behind S at distance d. By how much is IP multiplied by the presence of the mirrorExplanation / Answer
Since the object is 4.5 d in the front of the mirror therefore virtual object will be 4.75 d behind the mirror
so the distance between the virtual object and the screen is
(4.75 +4.75 +1 )d = 10.5d
Hence the intensity become
= 1+ (d/10.5d)2 = 1.009 times