Unit 14: Waves , Sound , and Physical Optics
Weightage : 15–18%
A light ray is incident normal to a thin layer of glass. Given the figure, what is the minimum thickness of the glass that gives the reflected light an orange-like color? The wavelength of orange light in air is \(600\,\mathrm{nm}\).
Correct Answer: \( \boxed{\mathrm{B}} \)
First, determine the wavelength of the light inside the glass using
\( n_1\lambda_1=n_2\lambda_2. \)
With \(n_{\mathrm{air}}=1.0\), \(n_{\mathrm{glass}}=1.5\), and \(\lambda_{\mathrm{air}}=600\,\mathrm{nm}\),
\( (1.0)(600)=(1.5)\lambda_{\mathrm{glass}}. \)
Therefore,
\( \lambda_{\mathrm{glass}}=\dfrac{600}{1.5}=400\,\mathrm{nm}. \)
At the air-glass boundary, the reflected ray undergoes a \(180^\circ\) (\(\tfrac{1}{2}\lambda\)) phase shift, while the reflection at the glass-water boundary undergoes no phase shift because the light reflects from a higher-index medium to a lower-index medium.
For constructive interference of the reflected light, the additional path through the film must contribute another \(\tfrac{1}{2}\lambda\) phase difference. This occurs for the minimum thickness
\( t=\dfrac{\lambda_{\mathrm{glass}}}{4}=\dfrac{400\,\mathrm{nm}}{4}=100\,\mathrm{nm}. \)
Hence, the minimum glass thickness is \(100\,\mathrm{nm}\), so the correct answer is (B).
A thin film with index of refraction \(n_f\) separates two materials, each of which has an index of refraction less than \(n_f\). A monochromatic beam of light is incident normally on the film, as shown above. If the light has wavelength \(\lambda\) within the film, maximum constructive interference between the incident beam and the reflected beam occurs for which of the following film thicknesses?
Correct Answer: \( \boxed{\mathrm{D}} \)
Since the film has a higher refractive index than both surrounding media,
\( n_{1}<n_{f} \) and \( n_{2}<n_{f} \),
the reflected ray at the top surface undergoes a phase change of \( \pi \) (equivalent to a path difference of \( \dfrac{\lambda}{2} \)), while the reflected ray at the bottom surface undergoes no phase change.
The light traveling through the film acquires an additional optical path difference of
\( 2t \),
where \(t\) is the film thickness and \(\lambda\) is the wavelength within the film.
Because there is one phase reversal, constructive interference in the reflected light occurs when
\( \displaystyle 2t=\left(m+\frac{1}{2}\right)\lambda,\qquad m=0,1,2,\ldots \)
The smallest nonzero thickness is obtained by taking \(m=0\):
\( \displaystyle t=\frac{\lambda}{4} \).
Therefore, the correct answer is (D).
A thin film of thickness \(t\) and index of refraction \(1.33\) coats a glass with index of refraction \(1.50\) as shown above. Which of the following thicknesses \(t\) will not reflect light with wavelength \(640\,\mathrm{nm}\) in air?
Correct Answer: \( \boxed{\mathrm{C}} \)
The wavelength of light inside the thin film is
\( n_{\mathrm{air}}\lambda_{\mathrm{air}}=n_{\mathrm{film}}\lambda_{\mathrm{film}} \)
\( (1.00)(640\,\mathrm{nm})=(1.33)\lambda_{\mathrm{film}} \)
\( \lambda_{\mathrm{film}}\approx\dfrac{640}{1.33}\approx481\,\mathrm{nm}. \)
At both interfaces (air \(\rightarrow\) film and film \(\rightarrow\) glass), the reflected ray undergoes a \(180^\circ\) phase shift because each reflection occurs from a lower to a higher refractive index. These two phase shifts cancel each other.
Therefore, to eliminate the reflected light (destructive interference), the optical path difference must be an odd multiple of
\( \dfrac{\lambda_{\mathrm{film}}}{2}, \)
which requires
\( t=\dfrac{(2m+1)\lambda_{\mathrm{film}}}{4}, \qquad m=0,1,2,\ldots \)
Possible thicknesses are approximately
\( \dfrac{\lambda_{\mathrm{film}}}{4}\approx120\,\mathrm{nm}, \qquad \dfrac{3\lambda_{\mathrm{film}}}{4}\approx361\,\mathrm{nm}, \qquad \dfrac{5\lambda_{\mathrm{film}}}{4}\approx601\,\mathrm{nm}. \)
Among the given choices, \(360\,\mathrm{nm}\) is approximately equal to \( \dfrac{3\lambda_{\mathrm{film}}}{4} \), producing destructive interference and therefore no reflected light.
Hence, the correct answer is (C).
