Question:
In a single silt diffraction with lambda= 500nm and lens of diameter 0.1 mm then width of central maxima , obtain on screen at a distance of 1m will be
a)5mm b) 1mm c)10mm d)2.5mm ( i need full explantion)
Answer:
Please recheck the problem for correctness of numerals and description. Usually, in single slit experiment, problem will not contain the lens of diameter but would contain the width of the slit. Hence, assuming 0.1 mm to be the width of the slit, let us proceed to solve.
a = width of the slit = 0.1 mm = 1 * 10-4 m
λ = wavelength of light = 500 nm = 500 * 10-9 m
D = distance of the screen = 1m
Width of the central fringe = λD / a = 500 * 10-9 * 1 / 1 * 10-4 = 500 * 10-5 = .5 * 10-2 m = 5mm
Not what you are looking for? Go ahead and submit the question, we will get back to you.
