Question

In an experiment with sodium light ( $5890 A^o$ ), an interference pattern is obtained in which $20$ equally spaced fringes occupy $2.30cm$ on the screen. On replacing the sodium light with another monochromatic source, $30$ fringes occupy $2.80cm$ on the screen. Find the wavelength of the light from the second source.

Answer 1

To solve this question, we first need to find the fringe width in both the scenarios using the given information. Then, we need to equate fringe width with the wavelength of the light source in each case. Finally, equating the two equations, we will get our answer.
$\beta = \dfrac{{total{\kern 1pt} {\kern 1pt} {\kern 1pt} width}}{{number{\kern 1pt} {\kern 1pt} {\kern 1pt} of{\kern 1pt} fringes}}$ where $\beta$ is the fringe width
$\beta = \dfrac{{\lambda D}}{d}$ where $\beta$ is the fringe width, $D$ is the distance between screen and slits and $d$ is the distance between the slits.

Complete answer:
According to the information given in the question
Case $1$
number{\kern 1pt} {\kern 1pt} {\kern 1pt} of{\kern 1pt} fringes = 20 \\\ space{\kern 1pt} {\kern 1pt} {\kern 1pt} occupied = 2.30cm \\\
For light of wavelength ${\lambda _1} = 5890{\AA}$

Case $2$
number{\kern 1pt} {\kern 1pt} {\kern 1pt} of{\kern 1pt} fringes = 30 \\\ space{\kern 1pt} {\kern 1pt} {\kern 1pt} occupied = 2.80cm \\\
For light of wavelength ${\lambda _2}$
Now, using the eq $\beta = \dfrac{{total{\kern 1pt} {\kern 1pt} {\kern 1pt} width}}{{number{\kern 1pt} {\kern 1pt} {\kern 1pt} of{\kern 1pt} fringes}}$ for case $1$ and $2$ , we get
${\beta _1} = \dfrac{{2.3}}{{20}}$ ------(i)
${\beta _2} = \dfrac{{2.8}}{{30}}$ -------(ii)
Now, from the equation $\beta = \dfrac{{\lambda D}}{d}$ , where $D$ and $d$ are unchanged in both the cases, we get the equations
${\beta _1} = \dfrac{{{\lambda _1}D}}{d}$ ---(iii)
${\beta _2} = \dfrac{{{\lambda _2}D}}{d}$ ---(iv)
Then, dividing eq (iii) with eq (iv) , we get
$\Rightarrow \dfrac{{{\beta _1}}}{{{\beta _2}}} = \dfrac{{\dfrac{{{\lambda _1}D}}{d}}}{{\dfrac{{{\lambda _2}D}}{d}}} \\\ \Rightarrow \dfrac{{{\beta _1}}}{{{\beta _2}}} = \dfrac{{{\lambda _1}}}{{{\lambda _2}}} \\\$
Now, substituting the respective values in their respective places, we get
$\Rightarrow \dfrac{{\dfrac{{2.3}}{{20}}}}{{\dfrac{{2.8}}{{30}}}} = \dfrac{{5890}}{{{\lambda _2}}} \\\ \Rightarrow \dfrac{{2.3}}{{20}} \times \dfrac{{30}}{{2.8}} = \dfrac{{5890}}{{{\lambda _2}}} \\\ \Rightarrow \dfrac{{2.3}}{{2.8}} \times \dfrac{3}{2} = \dfrac{{5890}}{{{\lambda _2}}} \\\$
Now, switching the numerator and denominator on both sides, we have
$\Rightarrow \dfrac{{2.8}}{{2.3}} \times \dfrac{2}{3} = \dfrac{{{\lambda _2}}}{{5890}} \\\ \Rightarrow \dfrac{{{\lambda _2}}}{{5890}} = \dfrac{{2.8}}{{2.3}} \times \dfrac{2}{3} \\\ \Rightarrow {\lambda _2} = 5890 \times \dfrac{{2.8}}{{2.3}} \times \dfrac{2}{3} \\\ \Rightarrow {\lambda _2} = 4780{A^o} \\\$
Therefore, the wavelength of the light from the second source is $4780{A^o}$ .

Note:
In the given question, it is specified that only the source of light is changed. So, we infer that $D$ and $d$ remain unchanged and hence cancel each other out. Also, the space occupied by the fringes can be taken in $mm$ or $cm$ both as they cancel out each other. But, in case the values of $D$ and $d$ are specified, make sure to convert all of them either in $mm$ or $cm$ .