Question

Light of wavelength \(6000 \, \text{Å}\) is incident on a thin glass plate of refractive index \(1.5\) such that the angle of refraction into the plate is \(60^\circ\). Calculate the smallest thickness of the plate which will make a dark fringe by reflected beam interference.

• A. \(1.5 \times 10^{-7} \, \text{m}\)
• B. \(2 \times 10^{-7} \, \text{m}\)
• C. \(3.5 \times 10^{-7} \, \text{m}\)
• D. \(4 \times 10^{-7} \, \text{m}\)

Hint:

Use the thin-film interference formula to calculate the thickness of the film for dark fringes, keeping in mind the refractive index and the angle of refraction.

Answer 1

The Correct Option is D

For a dark fringe in reflected light, the condition is:

\[ 2\mu t \cos r = \left(m + \frac{1}{2}\right)\lambda, \]

where:

• \(\mu = 1.5\) (refractive index),
• \(t\) (plate thickness),
• \(r = 60^\circ\) (angle of refraction),
• \(\lambda = 6000 \, \text{\AA} = 6 \times 10^{-7} \, \text{m}\) (wavelength),
• \(m = 0\) (smallest thickness).

With \(m = 0\):

\[ 2\mu t \cos r = \frac{\lambda}{2}. \]

Solving for \(t\):

\[ t = \frac{\lambda}{4\mu \cos r}. \]

Substituting values:

\[ t = \frac{6 \times 10^{-7}}{4 \cdot 1.5 \cdot \cos 60^\circ}. \]

Since \(\cos 60^\circ = \frac{1}{2}\):

\[ t = \frac{6 \times 10^{-7}}{4 \cdot 1.5 \cdot \frac{1}{2}} = \frac{6 \times 10^{-7}}{3} = 2 \times 10^{-7} \, \text{m}. \]

Therefore:

\[ t = 2 \times 10^{-7} \, \text{m}. \]

Conclusion: The smallest thickness for a dark fringe is:

\[ 2 \times 10^{-7} \, \text{m}. \]