JEE MAIN - Physics (2020 - 8th January Evening Slot - No. 18)
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 :
0.853
0.568
0.672
0.760
Explanation
$$\Delta $$X = $${\lambda \over 8}$$
$$\Delta $$$$\phi $$ = $${{2\pi } \over \lambda }$$$$\Delta $$X = $${{2\pi } \over \lambda }{\lambda \over 8}$$ = $${\pi \over 4}$$
I = I0 $${\cos ^2}\left( {{{\Delta \phi } \over 2}} \right)$$
$$ \Rightarrow $$ $${I \over {{I_0}}}$$ = $${\cos ^2}\left( {{{{\pi \over 4}} \over 2}} \right)$$
$$ \Rightarrow $$ $${I \over {{I_0}}}$$ = $${\cos ^2}\left( {{\pi \over 8}} \right)$$ = 0.853
$$\Delta $$$$\phi $$ = $${{2\pi } \over \lambda }$$$$\Delta $$X = $${{2\pi } \over \lambda }{\lambda \over 8}$$ = $${\pi \over 4}$$
I = I0 $${\cos ^2}\left( {{{\Delta \phi } \over 2}} \right)$$
$$ \Rightarrow $$ $${I \over {{I_0}}}$$ = $${\cos ^2}\left( {{{{\pi \over 4}} \over 2}} \right)$$
$$ \Rightarrow $$ $${I \over {{I_0}}}$$ = $${\cos ^2}\left( {{\pi \over 8}} \right)$$ = 0.853
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