In a Young's double slit experiment the intensity at a point where the path difference is  λ6(λ being the wavelength of the light used) is I .If Io denotes the maximum intensity II0is equal to

The resultant intensity at a point where the path difference is x is given by IRES= 4 I0COS2πXλ The resultant intensity is given byI=I1+I2+2I1I2cos⁡δ Iresultant =I+I+2Icos⁡(π/3) ∵δ=2πλΔx=2πλ×16=π3 Iresultant =3I Imax=4I  Therefore, Iresultant Imax=34