Question 131: Waves Optics

Question

In the figure below, P and Q are two equally intense coherent sources emitting radiation of
wavelength 20 m. The separation between P and Q is 5 m and the phase of P is ahead of that of Q
by 90o. A, B and C are three distinct points of observation, each equidistant from the midpoint of
PQ. The intensities of radiation at A, B, C will be in the ratio :

Accepted Answer

Let the distace of each point from midpoint = d
For point A :
x
P
Ã¢ÂÂ x
Q
= (d + 2.5) Ã¢ÂÂ (d Ã¢ÂÂ 2.5) = 5 m
$$\Delta $$$$\phi $$ due to path difference = $${{2\pi } \over \lambda }\left( {\Delta x} ight)$$ = $${{2\pi } \over {20}}\left( 5 ight)$$ = $${\pi  \over 2}$$
At A, Q is ahead of P by path, as wave emitted by Q reaches before wave emitted by P.
Total phase difference at A
= $${\pi  \over 2} - {\pi  \over 2}$$ (due to P being ahead of Q by 90ÃÂº) = 0
$$ 	herefore $$ I
A
= I
1
+ I
2
+ $$2\sqrt {{I_1}} \sqrt {{I_2}} $$ cos $$\Delta $$$$\phi $$
= I + I + $$2\sqrt I \sqrt I \cos \left( 0 ight)$$
= 4I
For point C :
x
Q
Ã¢ÂÂ x
P
= (d + 2.5) Ã¢ÂÂ (d Ã¢ÂÂ 2.5) = 5 m
$$\Delta $$$$\phi $$ due to path difference = $${{2\pi } \over \lambda }\left( {\Delta x} ight)$$ = $${{2\pi } \over {20}}\left( 5 ight)$$ = $${\pi  \over 2}$$
Total phase difference at C = $${\pi  \over 2} + {\pi  \over 2}$$ = $$\pi $$
$$ 	herefore $$ I
C
= I
1
+ I
2
+ $$2\sqrt {{I_1}} \sqrt {{I_2}} $$ cos $$\Delta $$$$\phi $$
= I + I + $$2\sqrt I \sqrt I \cos \left( \pi ight)$$
= 0
For point B :
x
P
Ã¢ÂÂ x
Q
= 0
$$\Delta $$$$\phi $$ due to path difference = 0
$$\Delta $$$$\phi $$ = $${\pi  \over 2}$$ (Due to P being ahead of Q by 90ÃÂº)
$$ 	herefore $$ I
B
= I
1
+ I
2
+ $$2\sqrt {{I_1}} \sqrt {{I_2}} $$ cos $$\Delta $$$$\phi $$
= I + I + $$2\sqrt I \sqrt I \cos \left( {{\pi  \over 2}} ight)$$
= 2I
I
A
: I
B
: I
C
= 4I : 2I : 0
= 2 : 1 : 0

Answer

4 : 1 : 0

Answer

2 : 1 : 0

Answer

0 : 1 : 2

Answer

0 : 1 : 4

Waves Optics

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