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In a double-slit experiment, the fourth-order maximum for a wavelength of 450 nm occurs at an angle of θ=90°. (a) What range of wavelengths in the visible range (400 nm to 700 nm) are not present in the third-order maxima? To eliminate all visible light in the fourth-order maximum, (b) should the slit separation be increased or decreased and (c) what least change is needed?

Short Answer

Expert verified

(a) The wavelength from 600nm to 700 nmwill be absent in the visible range.

(b) The slit separation for third order wavelength decreases.

(c) The least change needed in separation is 200nm.

Step by step solution

01

Identification of given data

The wavelength of light is λ4=450nm

The order of fringe for fourth order is m4=4

The angle for the fourth order fringe is θ=90°

The order of fringe for third order is m3=3

The phase difference of fringe pattern varies with the path difference. For minimum phase difference path difference should be minimum and vice versa.

02

Step 2(a): Determination of range of wavelengths not present in third order wavelength

The separation between slits is given as:

dsinθ=m4λ4

Substitute all the values in the above equation.

dsin90°=4450nmd=1800nm

The lowest wavelength absent in visible range is given as:

dsinθ=m3λ3

Substitute all the values in equation.

1800nmsin90°=3λ3λ3=600nm

The wavelength from 600nm to 700 nm will be absent in the visible range.

Therefore, the wavelength from 600nm to 700 nmwill be absent in the visible range.

03

Step 3(b): Whether slit separation increases or decreases

The slit separation varies inversely with wavelength of light. The wavelength of light for third order fringe increases so slit separation for third order wavelength of light decreases.

Therefore, the slit separation for third order wavelength decreases.

04

Step 4(c): Determination of path difference for maximum possible phase difference

The separation for lowest wavelength of visible range is given as:

dlsinθ=m4λl

Here, λl is the lowest wavelength of visible range and its value is 400nm.

Substitute all the values in equation.

dlsin90°=4400nmdl=1600nm

The least change needed is given as:

D=d-dl

Substitute all the values in equation.

D=1800nm-1600nmD=200nm

Therefore, the least change needed in separation is 200nm.

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