Question

A narrow slit is illuminated by a parallel beam of monochromatic light of wavelength λ equals to 6000 Å and the angular width of the central maxima in the resulting diffraction pattern is measured. When the slit is next illuminated by light of wavelength λ’, the angular width decreases by 30%. Calculate the value of the wavelength λ’.

Answer 1

d sin θ = λ (First diffraction)

In case of small angle,

sin θ =θ and dθ =λ

Then we can say half angular width will be,

θ = λ/d

Then,

Full- angular width will be,

w = 2θ = 2λ/d  

 w' = 2λ'/d

λ'/λ = w'/w

or λ' =λ (w'/w) = 6000 × 0.7 = 4200 Å

The wavelength of light will be 4200 Å as it is concluded from above solution.

Answer 2

angular width θ of central maxima and wavelength λ

are related as θ = λ/a , where a is slit width.

In first case we have,   a x θ = 6000 A°  .........................(1)

In second case, we have,  a x θ x (100-30)/100 = λ' ......................(2)

hence from (1) and (2), we get

λ' = a x θ x 0.7 = 6000 x 0.7

Hence, λ' = 4200 A°

CHEERS!!