JEE MAIN - Physics (2020 - 6th September Morning Slot - No. 8)
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 :
_6th_September_Morning_Slot_en_8_1.png)
4 : 1 : 0
2 : 1 : 0
0 : 1 : 2
0 : 1 : 4
Explanation
Let the distace of each point from midpoint = d
For point A :
xP – xQ = (d + 2.5) – (d – 2.5) = 5 m
$$\Delta $$$$\phi $$ due to path difference = $${{2\pi } \over \lambda }\left( {\Delta x} \right)$$ = $${{2\pi } \over {20}}\left( 5 \right)$$ = $${\pi \over 2}$$
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 90º) = 0
$$ \therefore $$ IA = I1 + I2 + $$2\sqrt {{I_1}} \sqrt {{I_2}} $$ cos $$\Delta $$$$\phi $$
= I + I + $$2\sqrt I \sqrt I \cos \left( 0 \right)$$
= 4I
For point C :
xQ – xP = (d + 2.5) – (d – 2.5) = 5 m
$$\Delta $$$$\phi $$ due to path difference = $${{2\pi } \over \lambda }\left( {\Delta x} \right)$$ = $${{2\pi } \over {20}}\left( 5 \right)$$ = $${\pi \over 2}$$
Total phase difference at C = $${\pi \over 2} + {\pi \over 2}$$ = $$\pi $$
$$ \therefore $$ IC = I1 + I2 + $$2\sqrt {{I_1}} \sqrt {{I_2}} $$ cos $$\Delta $$$$\phi $$
= I + I + $$2\sqrt I \sqrt I \cos \left( \pi \right)$$
= 0
For point B :
xP – xQ = 0
$$\Delta $$$$\phi $$ due to path difference = 0
$$\Delta $$$$\phi $$ = $${\pi \over 2}$$ (Due to P being ahead of Q by 90º)
$$ \therefore $$ IB = I1 + I2 + $$2\sqrt {{I_1}} \sqrt {{I_2}} $$ cos $$\Delta $$$$\phi $$
= I + I + $$2\sqrt I \sqrt I \cos \left( {{\pi \over 2}} \right)$$
= 2I
IA : IB : IC = 4I : 2I : 0
= 2 : 1 : 0
For point A :
xP – xQ = (d + 2.5) – (d – 2.5) = 5 m
$$\Delta $$$$\phi $$ due to path difference = $${{2\pi } \over \lambda }\left( {\Delta x} \right)$$ = $${{2\pi } \over {20}}\left( 5 \right)$$ = $${\pi \over 2}$$
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 90º) = 0
$$ \therefore $$ IA = I1 + I2 + $$2\sqrt {{I_1}} \sqrt {{I_2}} $$ cos $$\Delta $$$$\phi $$
= I + I + $$2\sqrt I \sqrt I \cos \left( 0 \right)$$
= 4I
For point C :
xQ – xP = (d + 2.5) – (d – 2.5) = 5 m
$$\Delta $$$$\phi $$ due to path difference = $${{2\pi } \over \lambda }\left( {\Delta x} \right)$$ = $${{2\pi } \over {20}}\left( 5 \right)$$ = $${\pi \over 2}$$
Total phase difference at C = $${\pi \over 2} + {\pi \over 2}$$ = $$\pi $$
$$ \therefore $$ IC = I1 + I2 + $$2\sqrt {{I_1}} \sqrt {{I_2}} $$ cos $$\Delta $$$$\phi $$
= I + I + $$2\sqrt I \sqrt I \cos \left( \pi \right)$$
= 0
For point B :
xP – xQ = 0
$$\Delta $$$$\phi $$ due to path difference = 0
$$\Delta $$$$\phi $$ = $${\pi \over 2}$$ (Due to P being ahead of Q by 90º)
$$ \therefore $$ IB = I1 + I2 + $$2\sqrt {{I_1}} \sqrt {{I_2}} $$ cos $$\Delta $$$$\phi $$
= I + I + $$2\sqrt I \sqrt I \cos \left( {{\pi \over 2}} \right)$$
= 2I
IA : IB : IC = 4I : 2I : 0
= 2 : 1 : 0
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