Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

The two light beams having intensities $I$ and $9I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\pi / 2$ at point $P$ and $\pi$ at point $Q$ . Then the difference between the resultant intensities at $\mathrm{P}$ and $\mathrm{Q}$ will be :
Option: 1
$2I$

Option: 2
$6I$

Option: 3
$5I$

Option: 4
$7I$

Answers (1)

best_answer

$\mathrm{I_{P}= I_{1}+I_{2}+2\sqrt{I_{1}}\sqrt{I_{2}}\cos \left ( \frac{\pi}{2} \right )}$
$\mathrm{I_{P}= 10\, I\: \rightarrow (1)}$

$\mathrm{I_{Q}= I+9I+2\sqrt{I}\sqrt{9I}\cos \left ( \pi \right )}$
$\mathrm{I_{Q}= 4I\: \rightarrow (2)}$

$\mathrm{I_{P}-I_{Q}= 6I}$

The correct option is (2)

Posted by

Sanket Gandhi

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

5 g of Na₂SO₄ was dissolved in x g of H₂O. The change in freezing point was found to be 3.82⁰C. If Na₂SO₄ is 81.5% ionised, the value of x (K

A capacitor is made of two square plates each of side 'a' making a very small angle $\alpha$

A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO₃ to give fraction A. The leftover organic phase was extracted with d

Trending Articles/News

anam-khan

JEE Main 2025: BTech CSE cut-offs for PwD candidates at NITs make entry easy; top institutes still selective

anam-khan

73 or 99 percentile? JEE Main result questioned, J-K candidate says, ‘Don’t create hype’

anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile

Latest Question