Answer:
distance between the two second-order minima is 2.8 cm
Explanation:
Given data
distance = 1.60 m
central maximum = 1.40 cm
first-order diffraction minima = 1.40 cm
to find out
distance between the two second-order minima
solution
we know that fringe width = first-order diffraction minima /2
fringe width = 1.40 /2 = 0.7 cm
and
we know fringe width of first order we calculate slit d
β1 = m1λD/d
d = m1λD/β1
and
fringe width of second order
β2 = m2λD/d
β2 = m2λD / ( m1λD/β1 )
β2 = ( m2 / m1 ) β1
we know the two first-order diffraction minima are separated by 1.40 cm
so
y = 2β2 = 2 ( m2 / m1 ) β1
put here value
y = 2 ( 2 / 1 ) 0.7
y = 2.8 cm
so distance between the two second-order minima is 2.8 cm
Don’t still need the answers or are u done and is it on edge
It is vital to take measurements at the same time everyday to determine the natural cycle, this will show the numbers on a regular basis. It is important because this will be basis of any action that needs to be taken. If it's not taken at the same time, it may disrupt the correct result or change the expected outcome.
A. iodine is located in period 5, group 17 and is classified
as a nonmetal
Group is the column, period is the raw.
