In a Young’s double slit experiment, 16 fringes are observed in a certain segment of the screen when light of a wavelength 700 nm is used. If the wavelength of light is changed to 400 nm, the number of fringes observed in the same segment of the screen would be
Solution
Let the length of segment is "$l$"
<br><br>Let N is the no. of fringes in "$l$"
and w is fringe width.
<br><br>$\therefore$ Nw = $l$
<br><br>$\Rightarrow$ N$\left( {{{\lambda D} \over d}} \right)$ = $l$
<br><br>As in both cases segment length is same.
<br><br>$\therefore$ ${{{{N_1}{\lambda _1}D} \over d} = l}$
<br><br> and ${{{{N_2}{\lambda _2}D} \over d} = l}$
<br><br>$\therefore$ ${{{{N_1}{\lambda _1}D} \over d}}$ = ${{{{N_2}{\lambda _2}D} \over d}}$
<br><br>$\Rightarrow$ ${{N_1}{\lambda _1}}$ = ${{N_2}{\lambda _2}}$
<br><br>$\Rightarrow$ 16 × 700 = N<sub>2</sub> × 400
<br><br>$\Rightarrow$ N<sub>2</sub> = 28
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
This question is part of PrepWiser's free JEE Main question bank. 197 more solved questions on Optics are available — start with the harder ones if your accuracy is >70%.