Sunlight is used in a double-slit interference experiment. The fourth-order maximum for a wavelength of 450 nm occurs at an angle of θ = 90°. Thus, it is on the verge of being eliminated from the pattern because θ cannot exceed 90° in Eq. 35-14. (a) What least wavelength in the visible range (400 nm to 700 nm) are not present in the third-order maxima? To eliminate all of the visible light in the fourth-order maximum, (b) what least change in separation is needed?
for the fourth order maxima m = 4wavelength λ = 450 nmθ = 90osince from theconstructive interferencedsinθ =m λd= m λ / Sin θ=4 x 450 / 1=1800 nma) if m = 3λ1 = d sinθ / m=4 x 450 / 3=600 nmwavelengthrange is 450 to600 nmb) since the slit separation d is related to λasdα λas λ increases only when d is alsoincreaseso d1 - d2 = (1/ Sin θ)(m1λ1 - m2λ2)=(1/Sin 90) (4 x 450-3 x 600)=0m