Light of wavelength 546 nm is allowed to illuminate the slits of Young's experiment. The separation between the slits is 0.10 mm and the distance of the screen form the slits where interference effects are observed is 20 cm. At what angle the first minimum will fall? What will be the linear distance on the screen between adjacent maxima? (Ans: 0.16°, 1.1 mm)




Data Given:


Wavelength of light = λ = 546 nm = 546 ｘ 10⁻⁹ m = 5.46 ｘ 10⁻⁷ m

Separation between the slits = d = 0.1 mm = 0.1 ｘ 10⁻³ m = 1.0 ｘ 10⁻⁴ m 

Distance between Slit and screen = L = 20 cm = 0.2 m

Order of 1st Minimum = m = 0


To Find:

Angle for 1st Minimum = θ = ?

Distance between the adjacent Maxima on the screen  = Δy =  ?


Solution:


Formula for `m^{th}` minima we have

d sin θ = ( m + `\frac {1}{2}`) Î»

putting the corresponding values

1.0 ｘ 10⁻⁴ m ï½˜ sin θ = ( 0 + `\frac {1}{2}`) ï½˜ 5.46 ｘ 10⁻⁷ m

1.0 ｘ 10⁻⁴ m ï½˜ sin θ = 2.73 ï½˜ 10⁻⁷ m

sin θ = `\frac {2.73 ｘ 10⁻⁷ m}{1.0 ｘ 10⁻⁴ m}`

sin θ = 2.73 ï½˜ 10⁻³

sin θ = 0.00273

θ = sin⁻¹ (0.00273)

θ = 0.156° = 0.16° -----------------Ans.


Now using formula for fringe spacing we have


Δy = `\frac {λ L}{d}`


by putting values


Δy =  `\frac {5.46 ｘ 10⁻⁷ m ｘ 0.2 m}{1.0 ｘ 10⁻⁴ m }`


Δy =  `\frac {1.092 ｘ 10⁻⁷ m^2}{1.0 ｘ 10⁻⁴ m }`


Δy =  1.092 ｘ 10⁻³ m 


Δy =  1.1 mm-------------Ans. (2)




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