n the figure, an isotropic point source of light S is positioned at distance d f
ID: 2041302 • Letter: N
Question
n 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 2.1d. By how much is IP multiplied by the presence of the mirror?
Chapter 34, Problem 003 Your answer is partially correct. Try again. is measured. Then a la em ror ? is placed behind S at In the gure, an isotropic point source of light s ?s positioned at distance d rom a viewing screen A and the light intensity rpat pont P (level with S distance 2.1d. By how much is Ip multiplied by the presence of the mirror? 2.1 d Numb.01 This answer has no units ? the tolerance is +/-296
Explanation / Answer
Intensity at a point 'x' distance away from an isotropic source = P / 4 pi x2
Hence initial intensity at P, Io = P/4 pi d2
With mirror placed on left , light also reaches P, after reflection from mirror. Total distance travelled by this light = (2.1 +3.1)d , to reach at P
Intensity at P due to reflected light = P/4 pi(5.2d)2 = Io/5.22
Total intensity at p = Io( 1+1/5.22)
= 1.037 Io
Hence Ip is multiplied by 1.037
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.