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Wave Optics

Question
CBSEENPH12038404

In a double-slit experiment, two coherent sources have slightly different intensities I and (I + δI), such that δ << I, show that resultant intensity at maxima is near 4I, while that at minima is nearly (δl)2/4l.

Solution

Given, Young's double slit experiment with two coherent sources.
Intensity of first source, I1 = I 
Intensity of second source, I2 = I + δI

Now, using the formula for resultant intensity,  

For maxima,

              I = I1+I2+2I1I2cos ϕ 

          Imax = I1+I2+2I1I2cos0°        = I+(I+δI)+2(I(I+δI) 

As δI<<I, therefore,  

          Imax = I+I+2I = 4I 

For minima, 

                   I = I1+I2+2I1I2 cos ϕ 

               Imin = I+(I+δI) + 2I(I+δI) cos 180° 

               Imin = 2I + δI - 2I1+δII1/2
                      = 2I + δI - 2I1+12δII+1212-12δII2
                            
                    = 2I + δI - 2I - δI + 14 I δII2
          Imin = δI24I. 

Hence proved.