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

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.