Question

A light source of wavelength \[{\lambda _1} = 400nm\] and \[{\lambda _2} = 600nm\] is used in a Young’s double slit experiment. If recorded fringe widths for \[{\lambda _1}\] and \[{\lambda _2}\] are \[{\beta _1}\] and \[{\beta _2}\], and the number of fringes for them within a distance \[y\] on one side of the central maximum are \[{m_1}\] and \[{m_2}\] respectively then
(A) \[{\beta _2} > {\beta _1}\]
(B) \[{m_1} > {m_2}\]
(C) From the central maximum, \[{3^{rd}}\] maximum of \[{\lambda _2}\] overlaps with \[{5^{th}}\] minimum of \[{\lambda _1}\]
(D) The angular separation of fringes for \[{\lambda _1}\] is greater than \[{\lambda _2}\]

Answer 1

Hint: The width of a fringe, and thus angular separation, is directly proportional to the wavelength of the light. For a given distance from central maximum the number of fringes is inversely proportional to fringe width.

Formula used:
\[\beta = \dfrac{{\lambda D}}{d}\], where \[D\] is the distance between the slit and screen, and \[d\] is the separation distance of the two slits.
\[y = m\beta \] where \[y\] is the distance of the bright fringes from the central maximum. \[y = \dfrac{{(2{m_1} - 1){\lambda _1}}}{2}\left( {\dfrac{D}{d}} \right)\] where \[y\] is location of the dark fringes from the central maximum.
\[\theta = \dfrac{{m\lambda }}{d}\] where \[\theta \] is the angular separation.

Complete step by step answer
To investigate option A, we recall the expressions for widths of the fringes, which are given by
\[{\beta _1} = \dfrac{{{\lambda _1}D}}{d}\] and \[{\beta _2} = \dfrac{{{\lambda _2}D}}{d}\]
From the value given in the question
\[{\lambda _2} > {\lambda _1}\],
Now, since \[D\] and \[d\] are constant and thus, equal for both wavelength, we can conclude that
\[{\beta _2} > {\beta _1}\]
Thus, option A is a solution.
To investigate option B, the formulas for the distance of a fringes from the central maximum are used and these are given by
\[y = {m_1}{\beta _1}\] and \[y = {m_2}{\beta _2}\]
\[y\] in the equation above are equal in both cases, since we’re considering the number of fringes for the same distance from the central maximum for both wavelengths.
Therefore,
\[{m_1}{\beta _1} = {m_2}{\beta _2}\]
Now since \[{\beta _2} > {\beta _1}\], for equality, \[{m_1} > {m_2}\].
Thus, option B is also correct.
For Option C, we must calculate the location of the \[{3^{rd}}\] maximum and the \[{5^{th}}\] minimum for equality.
The formula for maximum of \[{\lambda _2}\] is
\[y = {m_2}{\beta _2} = \dfrac{{{m_2}{\lambda _2}D}}{d}\]
Substituting the values and solving we get,
\[{y_3} = 3 \times 600\left( {\dfrac{D}{d}} \right) = 1800\dfrac{D}{d}nm\]
Now, the formula for minimum of \[{\lambda _1}\] is
\[y = \dfrac{{(2{m_1} - 1){\lambda _1}}}{2}\left( {\dfrac{D}{d}} \right)\]
Substituting the values and solving we get,
\[{y_5} = \dfrac{{(2(5) - 1) \times 400}}{2}\left( {\dfrac{D}{d}} \right) = 1800\dfrac{D}{d}nm\]
Therefore, the two fringes coincide. Thus, option C is correct.
For Option D, we write the formula for angular separation which is
\[\theta = \dfrac{{m\lambda }}{d}\]
Since \[{\lambda _2} > {\lambda _1}\], then \[{\theta _2} > {\theta _1}\] which does not match the option statement. Thus, option D is incorrect.
Hence, the correct options are A, B and C.

Note
In actual, the angular separation of fringes is given by \[\sin \theta = \dfrac{{m\lambda }}{d}\] but for small angles (\[\theta \ll 1\])\[\sin \theta \approx \theta \]. Thus, we have \[\theta = \dfrac{{m\lambda }}{d}\] where \[\theta \] is in radians. However, sometimes in reality, where high degree of accuracy is required and large angles are involved the more accurate form is used.