Question 138: Waves Optics

Question

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 :

Accepted Answer

$$\Delta $$X = $${\lambda  \over 8}$$
$$\Delta $$$$\phi $$ = $${{2\pi } \over \lambda }$$$$\Delta $$X = $${{2\pi } \over \lambda }{\lambda  \over 8}$$ = $${\pi  \over 4}$$
I = I
0
$${\cos ^2}\left( {{{\Delta \phi } \over 2}} \right)$$
$$ \Rightarrow $$ $${I \over {{I_0}}}$$ = $${\cos ^2}\left( {{{{\pi  \over 4}} \over 2}} \right)$$
$$ \Rightarrow $$ $${I \over {{I_0}}}$$ = $${\cos ^2}\left( {{\pi  \over 8}} \right)$$ = 0.853

Answer

0.853

Answer

0.568

Answer

0.672

Answer

0.760

Waves Optics

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 :

Choose the correct answer: