CBSE Class 12-science Answered
Young single slit
Asked by alanpeter9611 | 24 Feb, 2019, 12:26: AM
Expert Answer
(1) Intensity of diffraction fringes decreases as the order (n) increases
In the diffraction pattern of single slit, there are maxima at θ ≈ (n + 1/2) λ/a . Intensity is decreasing with increasing n.
Let us consider an angle θ = 3λ/2a which is midway between two of the dark fringes. Let us Divide the slit into three equal parts.
If we take the first two thirds of the slit, the path difference between the two ends would be
The first two-thirds of the slit can therefore be divided into two halves which have a path difference λ.
The contributions of these two halves cancel . Only the remaining one-third of the slit contributes to the intensity at a point between
the two minima. This will be much weaker than the central maximum (where the entire slit contributes in phase). One
can similarly show that there are maxima at (n + 1/2) θ/a with n = 2, 3, etc. These become weaker with increasing n,
since only one-fifth, one-seventh, etc., of the slit contributes in these cases.
can similarly show that there are maxima at (n + 1/2) θ/a with n = 2, 3, etc. These become weaker with increasing n,
since only one-fifth, one-seventh, etc., of the slit contributes in these cases.
(2) angular width of central maxima is twice that of the first order secondary maxima
First Dark fringe is occuring when path difference is λ. Hence angular width of central maxima = λa , where a is slit width.
Second dark fring is occuring when path difference is 2λ. Hence angular width of first order secondary maxima = 2λa
Answered by Thiyagarajan K | 24 Feb, 2019, 11:35: AM
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