JEE MAIN - Physics (2023 - 30th January Evening Shift - No. 20)
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
3
Explanation
$$I' = I{\cos ^2}\left( {{{k\Delta x} \over 2}} \right)$$
so $${I_1} = 4{I_0}{\cos ^2}\left( {{{2\pi } \over {2\lambda }} \times {\lambda \over 4}} \right)$$
$${I_1} = 2{I_0}$$
& $${I_2} = 4{I_0}{\cos ^2}\left( {{{2\pi } \over {2\lambda }} \times {\lambda \over 3}} \right)$$
$${I_2} = {I_0}$$
So $${{{I_1} + {I_2}} \over {{I_0}}} = 3$$
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