Question

A two-slit Young's interference experiment is arranged as illustrated (Fig.); l = 5000 Å. When a thin film of a transparent material is put behind one of the slits, the zero order fringe moves to the position previously occupied by the 4^th order bright fringe. The index of refraction of the film is n = 1.2. Calculate the thickness of the film.

• A. 5 µm
• B. 8 µm
• C. 10 µm
• D. 12 µm

Answer 1

Answer: (C)

10 µm

Answer 2

Intensity maxima occur when the optical path difference is D = ml.

Thus dD = ldm.

When the film is inserted as shown in Fig., the optical path changes bydD = t(n _ 1),

where t is the thickness of the film. As the interference pattern shifts by 4 fringes,

dm = 4.

Hence t(n – 1) = 4l, giving

t =$\frac{4\lambda}{n - 1}$= 10 µm