Question

A double-slit pattern is observed on a screen 2 m from the slits. Given that the light is incident normally and has a wavelength of 460 nm, what is the minimum slit separation for a point 3.2 mm from the center to be a) a minimum; or b) a maximum.

Answer 1

Use the condition for minima and maxima and use the distance of a certain band from the central band to find the minimum slit width for maxima and minima condition. For minima condition phase difference is zero while for maxima condition it is one.

Formula used:
The minima condition is given by, distance of band from central band,
$y = \left( {2m - 1} \right)\dfrac{{\lambda D}}{{2d}}$
where, $\lambda$ is the wavelength of the light $D$ is the distance of the screen from the slit and $m = 0,1,2,3...$.
The maxima condition is given by, distance of band from central band,
$y = \dfrac{{m\lambda D}}{d}$
where, $d$ is the separation of slits $D$ is the screen distance.

Complete step by step answer:
We know that that the distance from central band for a double slit is given by, $y = \left( {2m - 1} \right)\dfrac{{\lambda D}}{{2d}}$ for minima and $y = \dfrac{{m\lambda D}}{d}$ for maxima where, $\lambda$ is the wavelength of the light, $d$ is the separation of slits $D$ is the screen distance.

(a) Now, the slit width will be a minimum if the band forming in the spectrum is of the lowest order. So, $m$ will be 1. Hence, for minimum slit width for minima at 3.2mm,
$y = \left( {2.1 - 1} \right)\dfrac{{\lambda D}}{{2d}}$
$\Rightarrow d = \dfrac{{\lambda D}}{{2y}}$
Putting the values, $\lambda = 460\,nm = 460 \times {10^{ - 9}}m$, $D = 2\,m$, $y = 3.2\,mm = 3.2 \times {10^{ - 3}}m$, we will have,
$d = \dfrac{{460 \times {{10}^{ - 9}} \times 2}}{{3.2 \times {{10}^{ - 3}} \times 2}}$
$\therefore d = 143.75 \times {10^{ - 6}}\,m$

(b) So, for minimum slit width for maxima at 3.2mm,
$y = \dfrac{{\lambda D}}{d}$
$\Rightarrow d = \dfrac{{m\lambda D}}{y}$
Putting the values,$\lambda = 460\,nm = 460 \times {10^{ - 9}}m$, $D = 2\,m$, $y = 3.2\,mm = 3.2 \times {10^{ - 3}}m$, we will have,
$d = \dfrac{{460 \times {{10}^{ - 9}} \times 2}}{{3.2 \times {{10}^{ - 3}}}}$
$\therefore d = 287.5 \times {10^{ - 6}}\,m$

Hence, the minimum slit width at a) a minimum is $143.75 \times {10^{ - 6}}m$ and at (b) a maximum is $287.5 \times {10^{ - 6}}m$.

Note: When putting the values do not forget to convert the values into SI units, else the answer will be incorrect. The order of the band should be a minimum to have the slit width a minimum separation. Note that the slit width is halved for minimum condition. Since, the value of d is dependent on the value of $n$.