JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 16)
In a Young's double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is $${1 \over 8}$$ th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to :
0.94
0.85
0.74
0.80
설명
$$\Delta $$x $$=$$ $${\lambda \over 8}$$
$$\Delta $$$$\phi $$ $$=$$ $${{\left( {2\pi } \right)} \over \lambda }{\lambda \over 8} = {\pi \over 4}$$
I $$=$$ I0cos2$$\left( {{\pi \over 8}} \right)$$
$${{\rm I} \over {{{\rm I}_0}}} = $$ cos2$$\left( {{\pi \over 8}} \right)$$
$$\Delta $$$$\phi $$ $$=$$ $${{\left( {2\pi } \right)} \over \lambda }{\lambda \over 8} = {\pi \over 4}$$
I $$=$$ I0cos2$$\left( {{\pi \over 8}} \right)$$
$${{\rm I} \over {{{\rm I}_0}}} = $$ cos2$$\left( {{\pi \over 8}} \right)$$