Urgent please.. why is there central bright fringe in newton's ring due to transmitted beam?

In the case of reflection,

the air-glass boundary causes a half-cycle phase shift because the air has a lower refractive index than the glass.

Condition for bright fringe is (2 μ t + λ/2) = n λ

Which reduces to 2 μ t = (2n-1) λ/2

Condition for dark fringe is (2 μ t + λ/2) = (n + 1) λ/2

Which reduces to 2 μ t = n λ

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In the case of transmitted beam there is no such additional path difference

Path difference is 2 t

For bright fringe

2t = n λ

For dark fringe

(2 t) = (n- 1) λ/2 { for n = 1 no path difference , for n= 2 there is a path difference of λ/2 }

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We can get bright and dark fringe in both reflected and transmitted part of lens.

The path difference for bright fringe = n*lamda and for dark fringe (2n-1)lambda/2 This equations comes from derivation.

we get imperative dark fringe while there is no hollow between the convex and flat lenses. we get imperative mind-blowing fringe while there is minute hollow or clearence between the convex and flat lenses. answer