Question

In a single slit diffraction experiment, a slit of width 'd' is illuminated by red light of wavelength 650 nm. For what value of 'd' will
(i)    the first minimum fall at an angle of diffraction of 30°, and
(ii)    the first maximum fall at an angle of diffraction of 30°?

Answer 1

The first minimum fall occurs at 1300nm and first maximum fall occurs at 1950nm, in a single slit diffraction experiment, a slit of width 'd' is illuminated by red light of wavelength 650 nm.

Let wavelength, L = 650nm = 650 x 10^-9 m .

let x be the angle of diffraction.

for minimum fall, nL = dsinx &

for maximum fall, (2n +1)L/2 = dsinx

(i) for first minimum fall, n =1, x = 30degree

    L = dsinx ==> 650nm = d x sin 30;

   d = 650nm/sin 30 = 1300nm

(ii) for first maximum fall, n=1, x = 30 degree

   (2+1)L/2 = d sin 30 ==> 3X650nm = d x sin 30

  d = 3XL X2/2 = 3X650nm = 1950nm

Answer 2

I)The position of maximum is given as,

(2n+1)

2

λ

=asinθ

3×650=2asin30

∘

a=1950nm

a=1.95μm

Thus, for first maximum fall the value of a is 1.95μm.