Question

In Young’s double slit experiment when sodium light of wavelength $5893\overset{\circ }{\mathop{A}}\,$ is used, then 62 fringes are seen in the field of view. Instead, if violet light of wavelength $4358\overset{\circ }{\mathop{A}}\,$ is used, then the number of fringes that will be seen in the field of view will bee
A. 54
B. 64
C. 74
D. 84

Answer 1

Hint: Since the field of view is the same in both the cases, the distance on the screen of the view will be the same in both the cases. Write down the expressions for the fringes widths in both cases and by using the formula for the distance taken by n fringes find the number of fringes in the second case.

Formula used:
$y=n\beta $
where y is the distance on the screen covered by n fringes with $\beta $ as the fringes width.
 $\beta =\dfrac{D\lambda }{d}$
where $\lambda $ is the wavelength of the light, D is the distance between the slits and the screen and d is the distance between the two slits.

Complete step by step answer:
It is given that the in first case, sodium light of wavelength ${{\lambda }_{1}}=5893\overset{\circ }{\mathop{A}}\,$ is used in the Young’s double slit experiment. It is found that for a particular field of view, the number of fringes in this field of view are ${{n}_{1}}=62$.
Then, in the second violet light of wavelength ${{\lambda }_{2}}=4358\overset{\circ }{\mathop{A}}\,$ is used and for the same field of view, the fringes are observed. Let the number of fringes in the field of view for the violet light be ${{n}_{2}}$.
Since the field of view is same for both the case, the distance (height) of the screen for this field of view will be same in both cases.
And we know that $y=n\beta $.
In the first case, $y={{n}_{1}}{{\beta }_{1}}$ and in the second case $y={{n}_{2}}{{\beta }_{2}}$
With this we get that ${{n}_{1}}{{\beta }_{1}}={{n}_{2}}{{\beta }_{2}}$ …. (i)
And we know that ${{\beta }_{1}}=\dfrac{D{{\lambda }_{1}}}{d}$ and ${{\beta }_{2}}=\dfrac{D{{\lambda }_{2}}}{d}$.
Substitute these values in equation (i).
$\Rightarrow {{n}_{1}}\left( \dfrac{D{{\lambda }_{1}}}{d} \right)={{n}_{2}}\left( \dfrac{D{{\lambda }_{2}}}{d} \right)$
$\Rightarrow {{n}_{1}}{{\lambda }_{1}}={{n}_{2}}{{\lambda }_{2}}$ ….. (ii)
According to the given data in the question, we get that ${{n}_{1}}=62$, ${{\lambda }_{1}}=5893\overset{\circ }{\mathop{A}}\,$ and ${{\lambda }_{2}}=4358\overset{\circ }{\mathop{A}}\,$.
Now, substitute these values in equation (ii).
$\Rightarrow (62)(5893)={{n}_{2}}(4358)$
$\therefore {{n}_{2}}=\dfrac{(62)(5893)}{(4358)}=84$
This means that the number of fringes seen in the same field of view for the second case is 84.

Hence, the correct option D.

Note: when we say about fringes, they include both – bright fringes as well as dark fringes. Fringe width is the distance between any two consecutive dark fringes or two consecutive bright fringes.