In a double slit experiment, at a certain point on the screen the path difference between the two interfering waves is ${1 \over 8}$th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is :
Solution
$\Delta$X = ${\lambda \over 8}$
<br><br>$\Delta$$\phi$ = ${{2\pi } \over \lambda }$$\Delta$X = ${{2\pi } \over \lambda }{\lambda \over 8}$ = ${\pi \over 4}$
<br><br>I = I<sub>0</sub> ${\cos ^2}\left( {{{\Delta \phi } \over 2}} \right)$
<br><br>$\Rightarrow$ ${I \over {{I_0}}}$ = ${\cos ^2}\left( {{{{\pi \over 4}} \over 2}} \right)$
<br><br>$\Rightarrow$ ${I \over {{I_0}}}$ = ${\cos ^2}\left( {{\pi \over 8}} \right)$ = 0.853
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%.