Question

The distance of the 10th dark band from the centre of an interference pattern is $28.5mm$. The band width is-
(A) $3mm$
(B) $2.85mm$
(C) $4mm$
(D) $3.91mm$

Answer 1

We know that the band width is constant in an interference pattern. So, from the formula of the band width of a single band, by multiplying it with the number of bands, we can get the distance of the 10th band from the centre. From there, we get the band width of a single band.
Formula used: In this question, we will be using the following formula,
$\Rightarrow y = \dfrac{{n\lambda D}}{d}$
where $y$ is the distance from the centre and $n$ is the number of bands.
Here, the value $\dfrac{{\lambda D}}{d}$ gives the band width where
$\lambda$ is the wavelength of the wave and
$D$ is the distance of the screen from slits and $d$ is the distance between the slits.

Complete step by step answer:
In Young’s double-slit experiment, the source is placed behind the two slits such that an interference pattern is formed on the screen kept at a distance $D$ from the slits.
In the interference pattern, the distance between two consecutive dark bands or two consecutive bright bands is known as the band width. This band width remains constant for all the bands.
The value of this bandwidth is given by $\beta = \dfrac{{\lambda D}}{d}$ where $\beta$ is the band width.
Now the distance of the 10th dark band from the centre means that there is a distance of 10 times the band width between the centre and the 10th dark band.
Now, this distance is given by
$\Rightarrow y = \dfrac{{n\lambda D}}{d}$
or we can write,
$\Rightarrow y = n\beta$
In the question, we are given the values of $y$ and $n$ as
$n = 10$ and $y = 28.5mm$
So by substituting these values we get,
$\Rightarrow 28.5 = 10 \times \beta$
To find $\beta$ we take the 10 from the L.H.S to the R.H.S,
$\Rightarrow \beta = \dfrac{{28.5}}{{10}}$
On doing the calculation we have,
$\Rightarrow \beta = 2.85mm$
So the band width is $2.85mm$.
And the correct option is given by (B); $2.85mm$.

Note:
The distance between the two consecutive dark bands and the bright bands is always equal, as they are always equispaced from the central maxima. If white light is used in place of a monochromatic light then the white light breaks up into its components and we have the width of the red fringes greater than the blue fringes.