Question

In the young’s double slit experiment, the maximum number of bright bands observed (inclusive of the central bright band) is found to be $11$. If $\lambda $is the wavelength of the monochromatic light used, the distance between the slits is:
(A) $5\lambda $
(B) $6\lambda $
(C) $10\lambda $
(D) $11\lambda $

Answer 1

Hint
In this question, the distance between the slits can be calculated by multiplying the last number of Band (relative to the central bright fringe, considering it as one) with the wavelength of the monochromatic light and dividing it by the sine of the path difference.
$\Rightarrow d = \dfrac{{n\lambda }}{{\sin \theta }}$
Where d= distance between the two slits
n= number of fringe
$\theta $ angle made by the path difference of the waveform coming from two slits.
$\lambda $is the wavelength of the monochromatic light.

Complete step by step answer
We know that,
$\Rightarrow d = \dfrac{{n\lambda }}{{\sin \theta }}$
To find the value of$\sin \theta $,
As we know that a total of $11$ bright fringes is the maximum amount of fringes that can be produced.
It also includes the central bright fringe so, for the bright fringe n=1
Therefore, the number of nth fringe should be $5th$ (from the symmetry of the pattern, this will be the last possible fringe taken in one direction.
The angle $\theta $ can also be defined as the angle produced by the nth number of fringe.
And this is the maximum distance that the pattern could go, which pushes the $\sin \theta $ to its maximum value 1, which is for$\theta = 90^\circ $
Keeping these values in the formula we get,
$\Rightarrow d = \dfrac{{5\lambda }}{1}$
Therefore option (A) is the correct answer.

Note
The central fringe produced in the young’s double slit experiment is always bright so if there is a mention of ${p^{th}}$bright fringe, one should check if the central bright fringe is included or not, and if it is included, then the value of n must be $\dfrac{{p - 1}}{2}$. There is no such ambiguity for dark fringes.