In a Young's double slit experiment, the intensities at two points, for the path differences $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ( $\lambda$ being the wavelength of light used) are $I_{1}$ and $I_{2}$ respectively. If $I_{0}$ denotes the intensity produced by each one of the individual slits, then $\frac{I_{1}+I_{2}}{I_{0}}=$ __________.
Answer (integer)
3
Solution
<p>$I' = I{\cos ^2}\left( {{{k\Delta x} \over 2}} \right)$</p>
<p>so $${I_1} = 4{I_0}{\cos ^2}\left( {{{2\pi } \over {2\lambda }} \times {\lambda \over 4}} \right)$$</p>
<p>${I_1} = 2{I_0}$</p>
<p>& $${I_2} = 4{I_0}{\cos ^2}\left( {{{2\pi } \over {2\lambda }} \times {\lambda \over 3}} \right)$$</p>
<p>${I_2} = {I_0}$</p>
<p>So ${{{I_1} + {I_2}} \over {{I_0}}} = 3$</p>
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%.