Question

A double slit experiment is performed with light of wavelength $500\, nm$. A thin film of thickness $2\, \mu m$ and refractive index $1.5$ is introduced in the path of the upper beam. The location of the central maximum will

• A. remain unshifted
• B. shift downward by nearly two fringe
• C. shift upward by nearly two fringes
• D. shift downward by ten fringes

Answer 1

Answer: (B)

shift downward by nearly two fringe

Answer 2

When a thin film is placed in the path of one of the beams, then the optical path of that beam gets longer. The path difference is $\Delta x = (\mu - 1)t$ where $\mu$ is refractive index, t is thickness. Given $\mu = 1.5, t = 2 \times 10^{-6} \, m$ $\therefore ? x = (1.5 - 1) \times 2 \times 10^{-6} = 10^{-6} \, m = 1 \, \mu m$ Also path difference $? x = \frac{dy}{D}$ $\Rightarrow y = \frac{D}{d} ? x$ Also fringe width is given by $\beta = \frac{D \lambda}{d}$ $= \frac{D}{d} \times 500 \times 10^{-9} = \frac{D}{d} \times 0.5 \, \mu m$ $y = \frac{D}{d} \times 1 \, \mu m$ $= 2 \times \frac{D}{d} \times \frac{1}{2} \mu m =2\beta$ Hence, central maxima shifts upward by nearly two fringes.