Question

derive the expression for the intensity at a point where interference of light occurs arrive at the condition for maximum and zero intensity​

Answer 1

Answer:

Let y

1

and y

2

be displacement of two waves having same amplitude a and phase difference ϕ between them.

y

1

=asinωt

y

2

=asin(ωt+ϕ)

Resultant displacement is: y=y

1

+y

2

y=asinωt+asin(ωt+ϕ)=asinωt(1+cosϕ)+cosωt(asinϕ)

Rcosθ=a(1+cosϕ)

Rsinθ=asinϕ

y=Rsin(ωt+θ)

Where, R is resultant amplitude at P, I is intensity, squaring the equations we get,

I=R

2

=a

2

(1+cosϕ)

2

+a

2

(sinϕ)

2

=2a

2

(1+cosϕ)=4a

2

cos

2

2

ϕ

Maximum intensity:

cos

2

2

ϕ

=1

ϕ=2nπ where n=0,1,2,3,....

Therefore, I

max

=4a

2

Minmum intensity:

cos

2

2

ϕ

=0

ϕ=(2n+1)π where n=0,1,2,3,....

Therefore, I

max

=0

Explanation:

have a purplistic night

Answer 2

Answer:

Let y

1

and y

2

be displacement of two waves having same amplitude a and phase difference ϕ between them.

y

1

=asinωt

y

2

=asin(ωt+ϕ)

Resultant displacement is: y=y

1

+y

2

y=asinωt+asin(ωt+ϕ)=asinωt(1+cosϕ)+cosωt(asinϕ)

Rcosθ=a(1+cosϕ)

Rsinθ=asinϕ

y=Rsin(ωt+θ)

Where, R is resultant amplitude at P, I is intensity, squaring the equations we get,

I=R

2

=a

2

(1+cosϕ)

2

+a

2

(sinϕ)

2

=2a

2

(1+cosϕ)=4a

2

cos

2

2

ϕ

Maximum intensity:

cos

2

2

ϕ

=1

ϕ=2nπ where n=0,1,2,3,....

Therefore, I

max

=4a

2

Minmum intensity:

cos

2

2

ϕ

=0

ϕ=(2n+1)π where n=0,1,2,3,....

Therefore, I

max

=0

Explanation:

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