Optics 6 Question 34

36. A light source, which emits two wavelengths λ1=400nm and λ2=600nm, is used in a Young’s double-slit experiment. If recorded fringe widths for λ1 and λ2 are β1 and β2 and the number of fringes for them within a distance y on one side of the central maximum are m1 and m2, respectively, then

(a) β2>β1

(2014 Adv.)

(b) m1>m2

(c) from the central maximum, 3rd maximum of λ2 overlaps with 5 th minimum of λ1

(d) the angular separation of fringes of λ1 is greater than λ2

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Solution:

  1. Fringe width β=λDd

 or βλλ2>λ1 So β2>β1

Number of fringes in a given width

m=yβ or m1βm2<m1 as β2>β1

Distance of 3 rd maximum of λ2 from central maximum

=3λ2Dd=1800Dd

Distance of 5 th minimum of λ1 from central maximum

=9λ1D2d=1800Dd

So, 3 rd maximum of λ2 will overlap with 5 th minimum of λ1.

Angular separation (or angular fringe width) =λdλ

Angular separation for λ1 will be lesser.



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