In Young’s double slit experiment the intensity of light at a point on the screen where the path difference is λ is K…..

Q. In Young’s double slit experiment the intensity of light at a point on the screen where the path difference is λ  is K. The intensity of light at a point where the path difference is  λ /3 [ λ is the wavelength of light used] is

(a) K/4

(b) K/3

(c) K/2

(d) K

Ans:(a)

Sol: IR = I+I+2√(I.I)cosφ

IR = 2I(1+cosφ ) = 2I.2cos2(φ/2) = 4Icos2(φ/2)

A/c to question , K = 4Icosπ  (As path diff. is λ hence phase diff. is 2π)

K = 4I …….(i)

In 2nd case , IR‘ = 4Icos2(φ’/2)

IR‘ = 4Icos2(2π/3×2)   (As path diff. is λ/3 hence phase diff. is 2π/3)

= 4Icos2(π/3) = 4I×1/4 = I = K/4 from (i)

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