Question

A slit of width ‘d’ is illuminated by light of wavelength 5000 Å. For what value of ‘d’ will the first maximum fall at an angle of diffraction of 30˚.

Answer 1

Here, 

λ = 5000Å 

= 5× 10⁻⁷m 

d = ?; 

n = 1 ; 

θ = 30˚ 

For maxima of diffraction,

d sin θ = (2n + 1)\(\frac{λ}{2}\) 

or, d = \(\frac{(2n+1)\lambda}{2sin\theta}\) 

= \(\frac{3\times 5\times 10^{-7}}{2sin\,30°}\) 

= 1.5× 10⁻⁶ m.

Comments

Conditions for minima