Wavelength of Source in Loud Single Mirror Exp.

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In a loud single mirror experiment, a bright fringe of width 0.2 mm is observed at a distance of 1.0 m from the source, which changes to 0.5 mm when the mirror is raised by 0.9 mm. The wavelength of the source can be calculated using the formula that relates fringe width, distance between slits, and distance from the source to the screen. By applying the equation twice for both fringe widths, the distance between the slits is determined to be 1.5 mm, leading to a calculated wavelength of 6000 Å. The solution emphasizes the importance of unit consistency for grading or future similar problems. The final answer is confirmed as correct.
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1. Homework Statement : In a loud single mirror exp. a bright fringe of width 0.2 mm is seen at a point on the screen kept at a distance 1.0 m from the source. When the mirror is raised up to increase the angle of incidence at any point of the mirror by a dist. of 0.9 mm, fringe width of the same fringe becomes 0.5 mm, find the wavelength of the source.

2. Homework Equations : wavelenght = distance between the two slits (a) * fringe width (X) / distance from source to screen (D)
3. The Attempt at a Solution : How to find the distance between the two slits? Should I be proceeding by finding "a" by a= incidence angle*D?
 
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merc90 said:
1. Homework Statement : In a loud single mirror exp. a bright fringe of width 0.2 mm is seen at a point on the screen kept at a distance 1.0 m from the source. When the mirror is raised up to increase the angle of incidence at any point of the mirror by a dist. of 0.9 mm, fringe width of the same fringe becomes 0.5 mm, find the wavelength of the source.

2. Homework Equations : wavelenght = distance between the two slits (a) * fringe width (X) / distance from source to screen (D)

3. The Attempt at a Solution : How to find the distance between the two slits? Should I be proceeding by finding "a" by a= incidence angle*D?

Welcome to Physics Forums.

You can use the first equation you wrote,
wavelenght = distance between the two slits (a) * fringe width (X) / distance from source to screen (D)
... and write it twice, because there are two situations:
(1) the distance between the "slits" is a
(2) the distance between the "slits" is ____ [where you fill in the blank here].
 
Thanks Redbelly98 for the quick reply.

Is the answer: wavelength = 6000 A correct?

Steps:
wavelength = 2*a*X/D
Therefore, 2*a*0.2/1000=2*(a-0.9)*0.5/1000
which gives, a=1.5 or 2*a = 3.0
so, wavelength =3.0*0.2/1000
=6000 A
 
Looks good!

You may need to better explain how you worked the units, if you submit your homework to be graded or have to work a similar problem on an exam someday.
 
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