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In two-slit interference, if the slit separation is14μmand the slit widths are each 2.0μm, (a) how many two-slit maxima are in the central peak of the diffraction envelope and (b) how many are in either of the first side peak of the diffraction envelope?

Short Answer

Expert verified

(a) The range shows there will be 13 distinct values of m means 13 maxima’s in central envelope.

(b) There are 6 first side peaks in a diffraction envelope

Step by step solution

01

Identification of given data

The given data can be listed below,

  • The slit separation is,d=14μm
  • The slit width is,a=2.0mm
02

Concept/Significance of diffraction.

A wave bends in the direction of the obstacles when it travels through a tiny slit whose width is similar to the wave's wavelength. The screen's minima and maxima are then created by superimposing this twisted wave. The term for this occurrence is diffraction.

03

Determination of the two-slit maxima are in the central peak of the diffraction envelope

(a)

The initial diffraction minimum's angle,θ1=sin-1λa , is the range that the central diffraction envelope covers-θ1<θ<+θ1 -1 b +1. The double-slit pattern's maxima are located at is given by,

θm=sin-1mλd

So, the range of diffraction is given by,

-sin-1λa<sin-1mλd<+sin-1λa

Rearrange the range of the diffraction is given by,

-da<m<+da

The ratio of the slit separation and slit width is given by,

da=14μm2μm=7

Substitute all the values in the above,

-7<m<+7

Thus, the range shows there will be 13 distinct values of m means 13 maxima’s in central envelope.

04

Determination of number of the first side peak of the diffraction envelope

(b)

The range within one of the envelope is given by,

-sin-1λa<sin-1mλd<+sin-12λa

This results as-da<m<da gives 7<m<14

Since m is an integer, this indicates that8<m<13 and that there are 6 different values of m contained within that one envelope.

Thus, there are 6 first side peaks in a diffraction envelope.

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