Most Important Selected Qs for JEE AdvancedPhysicsWave Optics
In an interference arrangement, similar to Young's double-slit experiment, the slits S₁ and S₂ are illuminated with coherent microwave sources, each of frequency 10^6 ~Hz . The sources are synchronized to have zero phase difference. The slits are separated by distance d=150 m . The intensity I _ is measured as a function of , where is defined as shown in figure. If I₀ is maximum intensity, then I_ ( ) for 0 90^ is gi
Options
- AI_ ( ) =I₀ for =0^
- BI_ ( ) = (I₀ / 2 ) for =30^
- CI _ ( ) = ( I ₀ / 4 ) for =90^
- DI_ ( ) is constant for all values of
Correct answer
B. I_ ( ) = (I₀ / 2 ) for =30^
Step-by-step solution
For microwaves, c = f = c f = 3 10^8 10^6 =300 ~m Also, x = d , = 2 x = 2 ( ~d )= 2 300 (150 )= So, I=I₁+I₂+2 ( I₁ I₂ ) with I₁=I₂ and = , above equation reduces to I_R=2 I₁[1+ ( )]=4 I₁ ^2 ( 2 ) As I_R will be maximum when ^2[( ) / 2] is maximum, i.e., equal to 1 , so (I_R )_ =4 I₁=I₀ and hence I=I₀ ^2[( ) / 2] If =0^ , I=I₀ 0^ =I₀ If =30^ , I=I₀ ^2 ( 4 )= I₀ 2 If =90^ , I=I₀ ^2 ( 2 )=0