Question

Interference fringes are observed with a biprism of refracting angle 1 degree and refractive index 1.5 on a screen 100cm away from it .The wavelength of light used is 5890


a.If the distance between the source and the biprism is 20cm .The fringe width is:

Answer 1

given, n=1.5     r=20cm   D=100+20    A=10

                                         =120cm        =0.0174 radian

 

\delta =(n-1)A

   = (1.5-1)*0.0174

   =0.0087266

d=2r\delta

   = 2*20*0.0087266

   = 0.349 cm

B=D\lambda /d

     = (120*5890*10-8)/0.349

     = 0.02024 cm

     = 0.202 mm

Answer 2

Hello mate,

◆ Answer-
x = 0.2025 mm

◆ Explaination-
# Given-
λ = 5890 A = 5.89×10^-7 m
r = 20 cm = 0.2 m
D = 100+20 = 120 cm = 1.2 m
θ = 1° = 0.01745 rad
μ = 1.5

# Solution-
We know the formula -
δ = (μ-1)θ
δ = (1.5-1)0.01745
δ = 0.008725

Slit width can be calculated by -
d = 2rδ
d = 2 × 0.2 × 0.008725
d = 0.00349 m

Fringe width is calculated by formula -
x = λD / d
x = 5.89×10^-7 × 1.2 / 0.00349
x = 0.2025×10^-3 m
x = 0.2025 mm

Therefore, fringe width is 0.2025 mm.

Hope this helps you..