Please answer all the questions thoroughly and explain how you answered the ques
ID: 1302849 • Letter: P
Question
Please answer all the questions thoroughly and explain how you answered the question. Answer all the questions.
9. (A) Show that the spacing of two consecutive bright fringes is the same as that of two consecutive dark fringes.
(B) In a double-slit experiment using the He-Ne laser light of wavenlength 632.8 nm., the slit spacing is 0.05 mm. and the screen is 2m. from the slits. Find the distance along the screen between the third and fifth bright fringes.
10. The red monochromatic light from a He-Ne laser (632.8 nm.) is incident normally on a diffraction grating containing 6000 lines/mm. Find the angular separation between the first-order and second-order bright fringes. [Note: d=(1/6000) mm.]
Explanation / Answer
9(A)
Consider bright fringe.
y = mlL/d
For bright fringe m=1
y1 = (1)lL/d
for next order bright fringe m=2
y2 = (2) lL/d
fringe spacing = y2 - y1
or
Dx = (2)lL/d - (1)lL/d
Dx =lL/d (2-1)
Dx = lL/d
Similar result can be obtained for dark fringe.
(B)slit width = sw = 50000 [nm]
wavelength = wl = 632.8 [nm]
the angles of the minima (Am) are given by this equation:
sw * sin(Am) = n * wl
where n is 1,2,3,etc
sin(Am) = (n*wl)/sw
in this case:
"third bright fringe" indicates/implies...
n = 3
so now...
sin(Am) = (3*632.8)/50000
screen distance = sd = 2 [m]
distance on the screen from the central bright fringe to the third bright fringe above it
d1=tan(arcsin(sin(Am))) * sd =7.6 cm
for n=5
distance on the screen from the central bright fringe to the fifth bright fringe above it
d2=12.68 cm
The distance along the screen between the third and fifth bright fringes=d2-d1=5.08 cm
Fringe spacing or thickness of a dark fringe or a bright fringe is equal. It is denoted by Dx.
Consider bright fringe.
y = mlL/d
For bright fringe m=1
y1 = (1)lL/d
for next order bright fringe m=2
y2 = (2) lL/d
fringe spacing = y2 - y1
or
Dx = (2)lL/d - (1)lL/d
Dx =lL/d (2-1)
Dx = lL/d
Similar result can be obtained for dark fringe.
(B)slit width = sw = 50000 [nm]
wavelength = wl = 632.8 [nm]
the angles of the minima (Am) are given by this equation:
sw * sin(Am) = n * wl
where n is 1,2,3,etc
sin(Am) = (n*wl)/sw
in this case:
"third bright fringe" indicates/implies...
n = 3
so now...
sin(Am) = (3*632.8)/50000
screen distance = sd = 2 [m]
distance on the screen from the central bright fringe to the third bright fringe above it
d1=tan(arcsin(sin(Am))) * sd =7.6 cm
for n=5
distance on the screen from the central bright fringe to the fifth bright fringe above it
d2=12.68 cm
The distance along the screen between the third and fifth bright fringes=d2-d1=5.08 cm