In this question can we use the approach that
width of central maxima=2(distance of 5th maxima)
here I used 2(distance of 5th maxima as the question asks for 10 maximas.....so 5 above and 5 below.
Asked by aayush
23rd March 2018,
8:45 PM
Answered by Expert
Answer:
Yes. You can use as you mentioned in the posting of question.
In double slit experiment, slit spacing d = 1 mm , slits to screen distance D = 1 m, wavelength λ=500 nm.
fringe spacing = (Dλ)/d , this is worked out as 0.5 mm;
10 fringe spacing = 5 mm
In single slit experiment, width of central maximum is 2λ/a, where a is width of slit.
2λ/a = 5 mm , hence a = (2×500×10-9) / (5×10-3) = 0.2 mm
Answered by Expert
26th March 2018,
7:54 AM
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