Interference and Interferometers Questions and Answers

Q: If ‘\(A_{1}\)’ and ‘\(A_{2}\)’ are the amplitudes of two electromagnetic waves (same frequency) coming from two slits in Young’s double slit experiment, then the maximum intensity of interference fringe is :

The maximum intensity is \((A_{1} + A_{2})^{2}\).Intensity is proportional to the square of the resultant amplitude. With phase difference \(\delta\), the resultant amplitude A satisfies \(A^{2} = A_{1}^{2} + A_{2}^{2} + 2A_{1}A_{2}\cos\delta\). At a bright fringe \(\delta = 0\), so\[ \begin{aligned} I_{\max} &\propto A_{1}^{2} + A_{2}^{2} + 2A_{1}A_{2} \\ &= (A_{1} + A_{2})^{2} \end{aligned} \]\(A_{1} + A_{2}\) is the maximum amplitude, not the intensity. \(A_{1}^{2} + A_{2}^{2}\) is the average intensity, what incoherent sources would give, and its square root is the corresponding amplitude.

Q: Two coherent light sources of intensities I and 9I are used in an interference experiment. The resultant intensity at points where the waves from the two sources superpose with a phase difference zero is :

The resultant intensity is 16I.For coherent sources, amplitudes add, and intensity goes as amplitude squared. The amplitudes are proportional to \(\sqrt{I}\) and \(\sqrt{9I} = 3\sqrt{I}\). With zero phase difference they add directly:\[ \begin{aligned} I_{R} &= \big(\sqrt{I} + \sqrt{9I}\big)^{2} \\ &= (1 + 3)^{2} I = 16I \end{aligned} \]Adding intensities instead would give 10I, which is the incoherent result and not an option. The options 9I and I are the individual intensities, and zero is not possible here: even with a phase difference of \(\pi\) the minimum is \((3 - 1)^{2}I = 4I\).

Q: Consider a source which emits radiation of 500 nm wavelength. The linewidth of the emitted radiation is 1 nm. The coherence length \( |l_{c}| \) is :

The coherence length is 250 μm.The coherence length is the distance over which the wave train stays in phase, \( l_{c} = \lambda^{2}/\Delta\lambda \):\[ \begin{aligned} l_{c} &= \frac{(500\ \text{nm})^{2}}{1\ \text{nm}} = 2.5 \times 10^{5}\ \text{nm} \\ &= 250\ \mu\text{m} \end{aligned} \]2·5 μm is off by a factor of 100, and 1·0 μm and 100 μm do not follow from the formula.

Q: When the movable mirror of Michelson interferometer is shifted through 0.0589 mm a shift of 200 fringes is observed ? The wavelength of light used in Å is :

The wavelength is 5890 Å.Moving the mirror by d changes the path by 2d, and each fringe corresponds to one wavelength of path change, so \(2d = N\lambda\):\[ \begin{aligned} \lambda &= \frac{2d}{N} = \frac{2 \times 0.0589\ \text{mm}}{200} \\ &= 5.89 \times 10^{-4}\ \text{mm} \end{aligned} \]This is \(5.89 \times 10^{-7}\) m, or 5890 Å, the sodium D line. The other values (5885, 5895, 6000) do not follow from the given shift and count.

Q: For a typical laboratory sizes and specification which of the following has better resolving power ?

The best resolving power belongs to the Fabry-Perot etalon (spelt Fabry-Perrot in the paper).The resolving power of a Fabry-Perot etalon is the order of interference times the effective number of interfering beams. With a plate separation of a few millimetres and high-reflectance coatings this reaches a million or more, enough to resolve hyperfine structure.A laboratory grating spectrometer reaches tens of thousands (the order times the number of lines), and a prism spectrometer or a constant deviation spectrometer, which is also a prism instrument, only a few thousand, limited by the base of the prism.

If ‘\(A_{1}\)’ and ‘\(A_{2}\)’ are the amplitudes of two electromagnetic waves (same frequency) coming from two slits in Young’s double slit experiment, then the maximum intensity of interference fringe is :

A.\(A_{1} + A_{2}\)
B.\(A_{1}^{2} + A_{2}^{2}\)
C.\((A_{1} + A_{2})^{2}\)
D.\(\sqrt{A_{1}^{2} + A_{2}^{2}}\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The maximum intensity is \((A_{1} + A_{2})^{2}\).

Intensity is proportional to the square of the resultant amplitude. With phase difference \(\delta\), the resultant amplitude A satisfies \(A^{2} = A_{1}^{2} + A_{2}^{2} + 2A_{1}A_{2}\cos\delta\). At a bright fringe \(\delta = 0\), so

\[ \begin{aligned} I_{\max} &\propto A_{1}^{2} + A_{2}^{2} + 2A_{1}A_{2} \\ &= (A_{1} + A_{2})^{2} \end{aligned} \]

\(A_{1} + A_{2}\) is the maximum amplitude, not the intensity. \(A_{1}^{2} + A_{2}^{2}\) is the average intensity, what incoherent sources would give, and its square root is the corresponding amplitude.

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Two coherent light sources of intensities I and 9I are used in an interference experiment. The resultant intensity at points where the waves from the two sources superpose with a phase difference zero is :

A.16I
B.9I
C.I
D.Zero

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The resultant intensity is 16I.

For coherent sources, amplitudes add, and intensity goes as amplitude squared. The amplitudes are proportional to \(\sqrt{I}\) and \(\sqrt{9I} = 3\sqrt{I}\). With zero phase difference they add directly:

\[ \begin{aligned} I_{R} &= \big(\sqrt{I} + \sqrt{9I}\big)^{2} \\ &= (1 + 3)^{2} I = 16I \end{aligned} \]

Adding intensities instead would give 10I, which is the incoherent result and not an option. The options 9I and I are the individual intensities, and zero is not possible here: even with a phase difference of \(\pi\) the minimum is \((3 - 1)^{2}I = 4I\).

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Consider a source which emits radiation of 500 nm wavelength. The linewidth of the emitted radiation is 1 nm. The coherence length \( |l_{c}| \) is :

A.2·5 μm
B.250 μm
C.1·0 μm
D.100 μm

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The coherence length is 250 μm.

The coherence length is the distance over which the wave train stays in phase, \( l_{c} = \lambda^{2}/\Delta\lambda \):

\[ \begin{aligned} l_{c} &= \frac{(500\ \text{nm})^{2}}{1\ \text{nm}} = 2.5 \times 10^{5}\ \text{nm} \\ &= 250\ \mu\text{m} \end{aligned} \]

2·5 μm is off by a factor of 100, and 1·0 μm and 100 μm do not follow from the formula.

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When the movable mirror of Michelson interferometer is shifted through 0.0589 mm a shift of 200 fringes is observed ? The wavelength of light used in Å is :

A.5885
B.5890
C.5895
D.6000

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The wavelength is 5890 Å.

Moving the mirror by d changes the path by 2d, and each fringe corresponds to one wavelength of path change, so \(2d = N\lambda\):

\[ \begin{aligned} \lambda &= \frac{2d}{N} = \frac{2 \times 0.0589\ \text{mm}}{200} \\ &= 5.89 \times 10^{-4}\ \text{mm} \end{aligned} \]

This is \(5.89 \times 10^{-7}\) m, or 5890 Å, the sodium D line. The other values (5885, 5895, 6000) do not follow from the given shift and count.

Discuss in Forum

For a typical laboratory sizes and specification which of the following has better resolving power ?

A.Prism spectrometer
B.Grating spectrometer
C.Fabry-Perrot etalon
D.Constant deviation spectrometer

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The best resolving power belongs to the Fabry-Perot etalon (spelt Fabry-Perrot in the paper).

The resolving power of a Fabry-Perot etalon is the order of interference times the effective number of interfering beams. With a plate separation of a few millimetres and high-reflectance coatings this reaches a million or more, enough to resolve hyperfine structure.

A laboratory grating spectrometer reaches tens of thousands (the order times the number of lines), and a prism spectrometer or a constant deviation spectrometer, which is also a prism instrument, only a few thousand, limited by the base of the prism.

Discuss in Forum

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