Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen- The phase difference between the beams is -2 at point Aand - at point B- Then- the difference between the resultant intensities at A and B is A- 2 IB- 4 IC- 5 ID- I

The correct options are B 4IAt point A nbsp; Resultant intensity I_A=I_1+I_2+2 √(I_1I_2 cosϕ) I_A=I_1+I_2=5I At point B I_B=I_1+I_2 2√(I_1 I_2) =I+4I 4I=I I_A ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Intensity Amplitude Relationship

Two beams of ...

Question

Two beams of light having intensities $I$ and $4 I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{π}{2}$ at point $A$ and $π$ at point $B$ . Then, the difference between the resultant intensities at $A$ and $B$ is

2 I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 I

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5 I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $4 I$
At point $A$
Resultant intensity
$I_{A} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2} cos ϕ}$
$I_{A} = I_{1} + I_{2} = 5 I$
At point $B$
$I_{B} = I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}}$
$= I + 4 I - 4 I = I$
$I_{A} - I_{B} = 5 I - I = 4 I$

Suggest Corrections

Similar questions

Q. Two beams of light having intensities

I

and

4 I

interfere to produce a fringe pattern on a screen. The phase difference between the beams is

\frac{π}{2}

at point

A

and

π

at point

B

. Then, the difference between the resultant intensities at

A

and

B

Q. Two beams of light having intensities

I

and

4 I

interfere to produce a fringe pattern on a screen. The phase difference between the beams is

\frac{π}{2}

at point

A

and

2 π

at point

B

. Then the difference between the resultant intensities at

B

and

A

will be

Q. Two beams of light having intensities I and 4I interferer to produce a fringe pattern on screen. The phase difference between the beam is

\frac{π}{2}

at point A and

2 π

at point B. Then find out the difference between the resultant intensities at A and B.

Q. Choose the Correct answer from alternative given.
Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is

\frac{π}{2}

at point A and

π

at point B. Then the difference between resultant intensities at A and B is

Q. Two beams of light having intensities

I

and

4 I

interfere to produce a fringe pattern on a screen. The phase difference between the beams is

\frac{π}{2}

at point

A

and

2 π

at point

B

. Then the difference between the resultant intensities at

B

and

A

will be

Join BYJU'S Learning Program

Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π2 at point A and π at point B. Then, the difference between the resultant intensities at A and B is

The correct option is B 4IAt point A Resultant intensity IA=I1+I2+2√I1I2cosϕ IA=I1+I2=5I At point B IB=I1+I2−2√I1I2 =I+4I−4I=I IA−IB=5I−I=4I

Two beams of light having intensities $I$ and $4 I$ interfere to produce a fringe pattern on a screen. The phase difference between the beams is $\frac{π}{2}$ at point $A$ and $π$ at point $B$ . Then, the difference between the resultant intensities at $A$ and $B$ is

The correct option is B $4 I$
At point $A$
Resultant intensity
$I_{A} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2} cos ϕ}$
$I_{A} = I_{1} + I_{2} = 5 I$
At point $B$
$I_{B} = I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}}$
$= I + 4 I - 4 I = I$
$I_{A} - I_{B} = 5 I - I = 4 I$