Q. The two coherent sources of equal intensity produce maximum intensity of 100 units at a point. If the intensity of one of the sources is reduced by 36% by reducing its width then the intensity of light at the same point will be
A) 90
B) 89
C) 67
D) 81
Explanation
(d) Intensity of each source $= I_{0} = \frac{100}{4} = 25unit$
If the intensity of one of the source is reduced by 36% then $I_{1} = 25\ unit$ and $I_{2} = 25 - 25 \times \frac{36}{100} = 16\ (unit)$
Hence resultant intensity at the same point will be
$I = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}} = 25 + 16 + 2\sqrt{25 \times 16} = 81unit$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.