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 destructive interference (dark fringe) to occur in reflected light, the condition is:

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

where:

• \(\mu = 1.5\) is the refractive index of the glass plate,
• \(t\) is the thickness of the plate,
• \(r = 60^\circ\) is the angle of refraction,
• \(\lambda = 6000 \, \text{\AA} = 6 \times 10^{-7} \, \text{m}\) is the wavelength of light in air,
• \(m = 0\) for the smallest thickness (first-order dark fringe).

Substitute \(m = 0\) into the formula:

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

Rearranging for \(t\):

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

Substitute the given values:

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

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

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

The thickness of the plate is:

\[ t = 4 \times 10^{-7} \, \text{m}. \]

Conclusion: The smallest thickness of the plate that results in a dark fringe by reflected beam interference is:

\[ 4 \times 10^{-7} \, \text{m}. \]

Answer 2

Light of wavelength $\lambda = 6000 \text{ Å} = 6000 \times 10^{-10} \text{ m}$ is incident on a thin glass plate of refractive index $\mu = 1.5$. The angle of refraction into the plate is $r = 60^\circ$. We need to find the smallest thickness $t$ of the plate for a dark fringe in the reflected beam.

The condition for dark fringe in the reflected light from a thin film, considering the $\pi$ phase change at one reflection, is given by the hint:

$2 \mu t \cos r = n \lambda$, where $n = 1, 2, 3, ...$

For the smallest thickness, we take the smallest positive integer value for $n$, which is $n = 1$.

$2 \mu t \cos r = \lambda$

Now, we solve for the thickness $t$:

$t = \frac{\lambda}{2 \mu \cos r}$

Substitute the given values:

$\lambda = 6000 \times 10^{-10} \text{ m}$

$\mu = 1.5$

$r = 60^\circ$, so $\cos r = \cos 60^\circ = \frac{1}{2}$

$t = \frac{6000 \times 10^{-10}}{2 \times 1.5 \times \frac{1}{2}}$

$t = \frac{6000 \times 10^{-10}}{3 \times \frac{1}{2}}$

$t = \frac{6000 \times 10^{-10}}{1.5}$

$t = 4000 \times 10^{-10} \text{ m}$

$t = 4 \times 10^{-7} \text{ m}$

This corresponds to option (D).