Question

What should be the width of each slit to obtain the 10th maxima of the double slit pattern within the central maxima of the single slit pattern?

Answer 1

Hint Using young’s double slit experiment for bright brings of maxima is, $\Rightarrow$ $x=\dfrac{n\lambda D}{d}$
N = 0, 1, 2……D is distance of slit from screen and use angular separation between n maxima is, given by, $Q=\dfrac{n\lambda }{d}$using above formula, we can find the width of each slit.

Complete Step by Step Solution
Let width of each slit = a
And Distance between the slits is given by d.
In double slit experiment, separation between n maxima is given by
Xn = $\dfrac{n\lambda D}{d}$
Angular separation between n maxima is given by
${{\theta }_{n}}=\dfrac{n\lambda }{d}$
So, angular separation between 10 maxima,
${{\theta }_{10}}=\dfrac{10\lambda }{d}$------1
Path difference, $\lambda =asin$
$\lambda =a\theta$
$\theta =\dfrac{\lambda }{a}$
The angular width of central maxima in the diffraction pattern due to single slit of width
‘D’ is given by
$2\theta =2\left( \dfrac{\lambda }{a} \right)$------2
From eq 1 and 2
This is the expression for width of each slit
\begin{array}{*{35}{l}} {} & \left( \dfrac{10\lambda }{d} \right)=2\left( \dfrac{\lambda }{a} \right) \\\ {} & \left( \dfrac{5}{d} \right)=\dfrac{1}{a} \\\ {} & a=\dfrac{d}{5} \\\ \end{array}

Example If we take distance b/o the slits is 1mm, then d = 1mm
$a=\dfrac{1}{5}$= 0.2 mm

Note For drab fringes, path d1 Hence (minima)
$\begin{aligned} & \dfrac{xd}{D}=(2n-1)\dfrac{\lambda }{2} \\\ & x=(2n-1)\dfrac{\lambda D}{2d} \\\ \end{aligned}$
For dark brings or minima, use the above formula, for separation b/w two consecutive bright brings is the width of dark brings.
$\beta =\dfrac{\lambda D}{d}$
All bright and dark brings have equal width.