Question

derive the expression for intensity due to difraction of single slit. Find the condition of maxima and minima. Explain the intensity curve?​

Answer 1

Explanation:

derive the expression for intensity due to difraction of single slit. Find the condition of maxima and minima. Explain the intensity curve?

Answer 2

Answer:

Answer is 0

Explanation:

Let y1​ and y2​ be displacement of two waves having same amplitude a and phase difference ϕ between them.

y1​=asinωt

y2​=asin(ωt+ϕ)

Resultant displacement is: y=y1​+y2​

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=R2=a2(1+cosϕ)2+a2(sinϕ)2=2a2(1+cosϕ)=4a2cos22ϕ​

Maximum intensity:

cos22ϕ​=1

ϕ=2nπ where n=0,1,2,3,....

Therefore, Imax​=4a2

Minmum intensity:

cos22ϕ​=0

ϕ=(2n+1)π where n=0,1,2,3,....

Therefore, Imax​=0