Question

If in Young’s double-slit experiment, the slit width is 3cm, the separation between slits and screen is $70cm$ and wavelength of light is $1000\overset{\circ }{\mathop{A}}\,$ , then fringe width will be $\left( \mu =1.5 \right)$
(A) $2.3\times {{10}^{-3}}cm$
(B) $2.3\times {{10}^{-4}}m$
(C) $2.3\times {{10}^{-5}}cm$
(D) $2.3\times {{10}^{-6}}m$

Answer 1

Hint: Use the equation to find the ${{n}^{th}}$ order bright fringe and ${{\left( n+\dfrac{1}{2} \right)}^{th}}$ order dark fringe. Subtract both the values for bright and dark fringe for the same value $n$ . This subtracted value will give the fringe width in Young’s double slit experiment.

Complete step by step solution:
Let the fringe width be denoted by $\beta $ , slit width be represented by $d$ , the distance of separation of slit and screen be denoted by $D$ , and let the wavelength of light be denoted by $\lambda $ . Let $n$ denote the number of fringe under consideration. Let $y$ denote the position of bright or dark fringe.
The position of the ${{n}^{th}}$ bright fringe is given by
$y=\dfrac{nD\lambda }{d}$
The position for dark fringe is given by
$y=\left( n+\dfrac{1}{2} \right)\dfrac{D\lambda }{d}$
The difference between the two will give the fringe width:
$\beta =\dfrac{D\lambda }{d}$
Substituting the given values of $D$ ,$\lambda $ and $d$ , we get the fringe width as
$\beta =\dfrac{70\times {{10}^{-2}}\times 1000\times {{10}^{-10}}}{3\times {{10}^{-2}}}$
We have converted the units of distance to its SI unit.
Evaluating the above equation, we get
$\beta =2.34\times {{10}^{-6}}m$

Therefore, option (D) is the correct option.

Additional information:
Young’s double-slit experiment was first performed by Thomas Young in the year $1801$ . The experiment proves the wave nature of light. It involves the superposition of light to form a constructive interference and a destructive interference fringe pattern. It also shows the probabilistic behavior of a wave in quantum mechanics.

Note:
The SI unit of length is the meter. To convert angstrom to meter, we multiply by ${{10}^{-10}}$ .
It can be noted from the equation to find the fringe width that the fringe width directly depends on the distance of separation between the slits and the screen. So, the farther the distance, the bigger will be the fringe width.