Aditi Chauhan
Last Activity: 10 Years ago
Sol. Given that, t = 0.02 mm = 0.02 * 10^–3m, μ base 1 = 1.45, λ = 600 nm = 600 * 10^–9 m
a) Let, I base 1 = Intensity of source without paper = I
b) Then I base 2 = Intensity of source with paper = (4/9)I
⇒ l base 1/l base 2 = 9/4 ⇒ r base 1/r base 2 = 3/2 [because l ∝ r^2]
Where, r base 1 and r base 2 are corresponding amplitudes.
So, l base max/l base min = (r base 1 + r base 2)^2/(r base 1 - r base 2)^2 = 25 : 1
b) No. of fringes that will cross the origin is given by,
n = (μ – 1)/λ = (1.45 - 1) * 0.02 * 10^-3/600 * 10^-9 = 15.