Question
Download Solution PDFIn a Fresnel's biprism setup, monochromatic light of wavelength 550 nm is used to obtain interference fringes on the screen. On introducing a thin sheet of transparent material of thickness 1.1 microns in the path of one of the interfering beams, the central fringe is shifted through distance equal to the spacing between successive bright fringes. Then, the refractive index of the transparent material is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
When a thin plate of thickness t and refractive index μ is introduced in the path of one of the beams, then the entire fringe pattern gets shifted by a distance y towards the side on which the plate is placed.
The shifted distance y is given by
\(y = \frac{D}{d}\left( {μ - 1} \right)t\)
If Fresnel’s biprism experiment, the fringe width is given by
\(λ = \frac{{\beta d}}{D}\)
Combining the above equations, we get
\(y = \frac{\beta }{λ }\left( {μ - 1} \right)t\)
Calculation:
Given the central fringe is shifted through distance equal to the spacing between successive bright fringes ⇒ y = one fringe width = β
μ = 1.1, t = 1.1 × 10-6 m; λ = 550 nm = 550 × 10-9 m;
\(\beta = \frac{\beta }{λ }\left( {μ - 1} \right)t ⇒ λ = \left( {μ - 1} \right)t\)
⇒ 550 × 10-9 = (μ – 1) × 1.1 × 10-6 m
⇒ μ = 1 + 0.5 = 1.5
Last updated on Jun 19, 2025
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