y = – 4.33 mm Ι 0 = Ι max λ = 650 nm, 433.3 nm

Sol. Given –

λ = 600 mm = 6 × 10 -7 (in medium)

d = 0.45 mm = 0.45 × 10 -3 m

D = 1.5 m

Thickness of glass sheet, t = 10.4 μ m = 10.4 × 10 -6 m

Refractive index of medium sheet, μ m = 4/3

and refractive index of glass sheet, μ g = 1.5

Let central maximum is obtained at a distance y below point O. Then

Δ x 1 = S 1 P – S 2 P =

Path difference due to glass sheet

Δ x 2 = t

Net path difference will be zero when –

Δ x 1 = Δ x 2 or = t ∴ y = t

Substituting the values, we have

y = ⇒ y = 4.33 × 10 -3 m Ans.

At O, Δ x 1 = O and Δ x 2 = t

∴ Net path difference, Δ x = Δ x 2

Corresponding phase difference, Δφ or simply φ = . Δ x

Substituting the values, we have

φ = ( 10.4 × 10 -6 ) φ = π

Now I ( φ ) = I max cos 2

∴ I = I max cos 2 I = I max Ans.

At O : path difference is Δ x = Δ x 2 = t

For maximum intensity at O

Δ x = n λ ( Here n = 1, 2, 3......)

∴ λ = ............. and so on

λ = ( 10.4 × 10 6 m) = ( 10.4 × 10 3 nm )

λ = 1300 nm

∴ Maximum intensity will be corresponding to

λ = 1300 nm, nm, nm, nm.........

= 1300 nm, 650nm, 433nm.33nm, 325nm.......

The wavelengths in the range 400 to 700 nm are

650 nm and 433,33 nm