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Number of Fringes in a Diffraction Maximum. In Fig.36.12 c the central diffraction maximum contains exactly seven interference fringes, and in this case d =a =4. (a) What must the ratio d /a be if the central maximum contains exactly five fringes? (b) In the case considered in part (a), how many fringes are contained within the first diffraction maximum on one side of the central maximum?

Short Answer

Expert verified

a) The ratiod/a must be 3.0.

b) There are3.0 fringes contained with the first diffraction maximum on one side of the central maximum.

Step by step solution

01

Formula used to solve the question

Equation for the first minimum of the dark fringes:

asin=尘位

Intensity is given by

I=I0(cos22)(sin22)2

Where

=2蟺补sin

02

Calculate the ratio

Since the bright fringe contains5 fringes, so the minimum fringe that makes an interference of total dark is for m=3.

The equation is given by

asin=m

Hence,

asin= (1)

While it is the third bright fringe in the double-slit pattern,

dsin=3 (2)

Now, the intensity is given by

I=I0(cos22)(sin22)2

Where

=2蟺补sin

Thus,

I=I0(cos22)(sin蟺补蝉颈苍胃蟺补蝉颈苍胃)2

Now, for m=3, the intensity is zero.

0=I0(cos22)(sinasinasin)2sinasinasin=0sinasin=0

Plug the value,

sin(3asindsin)=03ad=1d/a=3.0

03

Calculate the number of fringes

Now, the fringe for the central maximum fringe are for

m=0,1,2

So,

asin=2 (3)

And

dsin=m (4)

Divide (4) by (3) and plug d /a,

dsinasin=m2m2=3m=6

This means that the second bright fringe envelope is contained betweenm=3 and m=6.

Implies that this envelope only contains two bright fringes which are for m=4,5.

Thus, the ratio d /a must be 3.0. There are3.0 fringes contained with the first diffraction maximum on one side of the central maximum.

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