Wavelength of Source in Loud Single Mirror Exp.

[merc90]

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?

 

[Redbelly98]

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].

 

[merc90]

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

 

[Redbelly98]

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.

 

[Nickie]

I would approach this problem by first understanding the experimental setup and the variables involved. In this case, we have a single mirror experiment where a source of light is reflected off the mirror and creates interference fringes on a screen placed at a distance of 1.0 m from the source. The fringes have a width of 0.2 mm.

When the mirror is raised by a distance of 0.9 mm, the angle of incidence at any point on the mirror increases, resulting in a change in the fringe width of the same fringe to 0.5 mm. To find the wavelength of the source, we can use the equation: wavelength = a * X / D, where a is the distance between the two slits, X is the fringe width, and D is the distance from the source to the screen.

To find the value of a, we can use the given information that the fringe width changed from 0.2 mm to 0.5 mm when the mirror was raised by 0.9 mm. This means that the distance between the two slits (a) is also 0.9 mm. Now, we can substitute the values in the equation to find the wavelength of the source.

Therefore, the wavelength of the source can be calculated as: wavelength = 0.9 mm * 0.5 mm / 1.0 m = 0.45 mm.

In conclusion, the wavelength of the source is 0.45 mm. It is important to note that this is an approximation as the values used in the calculation are rounded off. To get a more accurate value, the experiment can be repeated with more precise measurements.