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In YDSE with D = 1 m, d = 1 mm, light of...

In YDSE with `D = 1 m, d = 1 mm`, light of wavelength `500 nm` is incident at an angle of `0.57^(@)` w.r.t. the axis of symmetry of the experimental set-up. If center of symmetry of screen is O as shown figure.
a. find the position of centrla maxima,
b. find intensity at point O in terms of intensity of central maxima `I_(0)`, and
c. find number of maixma lying central maxima.
.

Text Solution

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(i) `theta=theta_(0)=0.57^(@)`
`rArr y=-D tan theta ~= -Dtheta=-1` meter `xx((0.57)/(57)rad) rArr y=-1cm.`
(ii) for point`0, theta=0`
Hence, `Deltap=d sin theta_(0), dtheta_(0)=1 mm xx(10^(-2)rad)`
`=10,000 nm=20xx(500 nm)`
`rArr Deltap=20 lambda`
Hence point `O,` corresponds to `20th` maxima `rArr` intensity at `O=I_(0)`
(iii) `19` maxima lie between central maxima and `O,` excluding maxima at `O` and central maxima.
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