CBSE Class 12-science Answered
please tell this question
Asked by rajeshbangia69 | 27 Mar, 2020, 01:16: PM
Expert Answer
Figure shows a lens with its central part of size 1 mm is made opaque.
S is source of light (λ = 500 nm ) is placed at 50 cm in front of lens.
Focal length of lens is 30 cm . Hence image is formed as two spots S1 and S2 at a distance v behind the lens as shown in figure.
we get distance v from lens equation, (1/v) - (1/u) = (1/f) [ By sign convention , lens-to-object distance u is negative ]
(1/v) = (1/30) - (1/50) or v = 75 cm .
Since the screen is placed at a distance 2 m from the lens, Screen is at (2-v) = (2-0.75) = 1.25 m from coherent sources S1 and S2
These two coherent sources S1 and S2 makes interference pattern on screen.
Distance between S1 and S2 is obtained using similar triangle properties.
( d / 1250) = 1/500 or d = 2.5 mm
Fringewidth β = ( D λ ) / d = ( 125 × 10-2 × 500 × 10-9 )/ (2.5 × 10-3) = 0.25 mm
highest order of fringe , nmax = d/λ = ( 2.5 × 10-3 )/(500×10-9 ) = 5000
Hence interference pattern is upto (5000×0.25 mm ) = 1.25 m on both side of point O on the screen
Minimum distance of screen from lens:- Screen can not be placed at a distance less than v = 75 cm ,
Screen should be placed at a distance greater than v = 75 cm.
Again fringe width β is proportional to D which is coherent sources to screen distance.
Hence minimum distance of screen from lens will be chosen by considering these two conditions.
Answered by Thiyagarajan K | 27 Mar, 2020, 06:20: PM
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