Question

Interference fringes are observed on a screen by illuminating two thin slits \[1mm\] apart with a light source ($\lambda = 632.8nm$). The distance between the screen and the slits is 100cm. If a bright fringe is observed on screen at a distance of $1.27mm$ from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to:
(A) $2.05\mu m$
(B) $2.87nm$
(C) $2nm$
(D) $1.27\mu m$

Answer 1

Hint: In order to solve this question, we will first find the order of the bright fringe obtained using general formula for fringes and then using formula of path difference of bright fringe we will solve for the value of path difference.

Formula used:
The order n for bright fringe is given by
$n = \dfrac {{yd}}{{D\lambda }}$
where y is the distance of fringe from central fringe, d is distance between slits, D is distance between slit and screen and $\lambda $ is the wavelength.

The path difference is given by
$\vartriangle x = n\lambda $

Complete answer:
We have given that, $d = 1mm = {10^{ - 3}}m$ , $\lambda = 632.8nm = 632.8 \times {10^{ - 9}}m$, $D = 100cm = 1m$ and $y = 1.27mm = 1.27 \times {10^{ - 3}}m$

On putting these values in $n = \dfrac{{yd}}{{D\lambda }}$ to find order we get,
$
  n = \dfrac{{(1.27 \times {{10}^{ - 3}})({{10}^{ - 3}})}}{{1(632.8 \times {{10}^{ - 9}})}} \\
  n = 2 \\
 $

So, the order of the bright fringe is $n = 2$ Now, put this value in $\vartriangle x = n\lambda $ and solve for path difference we get,
\[
  \vartriangle x = 2(632.8)nm \\
  \vartriangle x = 1265.6nm \\
  \vartriangle x = 1.27\mu m \\
 \]
So, the path difference is \[\vartriangle x = 1.27\mu m\]

Hence, the correct answer is option (D) $1.27\mu m$.

Note: It should be remembered that, while doing calculation in such problems always convert all the quantities unit in same units and here general conversion relations are used as
$
  1nm = {10^{ - 9}}m \\
  1\mu m = {10^{ - 6}}m \\
  1mm = {10^{ - 3}}m \\
 $
Hence, learn all the basic conversion metric units in order to avoid any calculation error in such type of problems.