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If the distance of a point on the screen
from the two sources is 1.8 x 10 m and
1.23 x 10-5 m and wavelength of light used is
6000 A then, it is
(a)
9th dark fringe
(b)
10th dark fringe
(c)
10th bright fringe
(d)
11th dark fringe
D
B
X₁
D
P(point)
Screen

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m, and d= 3.00 x 10 m. (B) Determine the distance between adjacent bright fringes. Use the bright fringe equation again to find the distance between any adjacent bright fringes (here, those characterized by m and m + 1). AL AL (m+ 1) - AL Ay = y+1 m%3D (5.63 x 107 m)(1.20 m) - 2.25 cm 3.00 x 105 m LEARN MORE REMARKS This calculation depends on the angle 0 being small because the small angle approximation was implicitly used. The measurement of the position of the bright fringes yields the wavelength of light, which in turn is a signature of atomic processes, as is discussed in the chapters on modern physics. This kind of measurement therefore helped open the world of the atom. QUESTION Which of the following make the separation between fringes greater in the two slit interference experiment? (Select all that apply.) O Larger separation of the two slits. O Narrower slits. Wider slits. O Smaller separation of the two slits. PRACTICE IT Use the worked example above to help you solve this…
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