In a Young’s experiment the sources are two slits of width ε = 1.0 µm, at 0.2 mm apart. They are illuminated by a mercury lamp of wavelength λ = 546 nm. The interference pattern is observed on a screen in the focal plane of a converging lens of focal length f = 100 cm. a) Determine the spacing

Question

In a Young’s experiment the sources are two slits of width ε = 1.0 µm, at 0.2 mm apart. They are illuminated by a mercury lamp of wavelength λ = 546 nm. The interference pattern is observed on a screen in the focal plane of a converging lens of focal length f = 100 cm. a) Determine the spacing between the fringes on this screen. What is the maximum number of observable fringes? What is the position of the fringe of order 100? b) What is the displacement of the fringes if we cover one of the slits with a transparent sheet of thickness e = 3 µm and an index of refraction n = 1.5?

Accepted Answer

a) The direction [latex] heta _p [/latex] of the fringe of order p is such that [latex] \sin heta _p=p\lambda /d [/latex] (Figure 8.4a). The fringe of order 0 is in the direction [latex] heta _o=0 [/latex] and the first fringe is in the direction [latex] heta _1 [/latex] such that [latex] heta _1=\lambda /d=2.73 imes 10^{-3} [/latex] . The spacing between fringes on the screen near O is [latex] i=f an heta _1=2.73 [/latex] mm. The angular width of diffraction is [latex] 2\Theta =2\lambda /ε [/latex]. This corresponds to a linear diffraction width 2λf/ε on the screen. Thus, the maximum number of observable fringes is [latex] N=2\lambda f/\varepsilon i=2d/\varepsilon =400 [/latex] .

The direction of the fringe of order 100 is such that [latex] \sin heta _{100}=100\lambda /d=0.273 [/latex] ;hence, [latex] \sin heta _{100}=p\lambda /d=15.84 . [/latex] This corresponds to a linear position on the screen [latex] x_{100}=f an heta _{100}=28.4 [/latex] cm ; this is slightly different from 100 i = 27.3 cm.

b) Let us assume that the sheet covers the lower slit [latex] S_2 [/latex] (Figure 8.4b). If light is observed in the direction θ, the angle of refraction in the sheet is [latex] heta ^\prime [/latex] such that [latex] \sin heta =n \sin heta ^\prime [/latex] . Light travels in the sheet a distance [latex] IJ=e/ \cos heta ^\prime [/latex] instead of [latex] IK=IJ \cos( heta - heta ^\prime)=e(\cos heta + an heta ^\prime \sin heta ). [/latex] Thus, we have to add an optical path difference to [latex] S_2H: \ nIJ-IK=ne/ \cos heta ^\prime-e(\cos heta + an heta ^\prime \sin heta )=e\left[\sqrt{n^2-\sin^2 heta }-\cos heta ight] \approx e(n-1), [/latex]

where we have assumed that the angles are small. This optical path difference modifies the order of interference by [latex] \delta p=(n-1)(e/\lambda ). [/latex] Thus, it displaces the fringes by i(n - 1)(e/λ) = 7.5 mm or more.

Question 8.1: In a Young’s experiment the sources are two slits of width ε...

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