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 resultant intensities at A and B is

A
5I
B
4I
C
3I
D
7I

Solution:

$I (ϕ) = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos ϕ H e r e, I_{1} = I a n d I_{2} = 4 I A t p o i n t A, ϕ = \frac{π}{2} ∴ I_{A} = I + 4 I + 0 = 5 I A t p o i n t B, ϕ = π ∴ I_{B} = I + 4 I - 4 I = I h e n c e . I_{A} - I_{B} = 4 I$

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

Solution:

Related content

Company

Support

Courses

More

JEE

NEET

CUET

CBSE

Popular books

NCERT solutions

NCERT exemplar

State Board

Subject

follow us