Question

A beam of light consisting of two wavelength 650nm and 520nm, is used to obtain interference fringes on a screen 1.2 m away. The separation between the slits is 2 mm.
What is the least distance from the central maximum when the bright fringes due to both the wavelength coincides?

Answer 1

nλ2 = (n-1) λ1           Where, λ1 = 650 nm

 

n= 5

 

 

Wavelength of another light beam (λ2) = 520 nm, Distance of the slits from the screen = D ,  Distance between the two slits = d

 

(a)

Distance of the nth bright fringe on the screen from the central maximum is given by the relation,

 

x = n λ1(D/d),  If third bright fringe. N=3

 

Then,

 

x = 3 x 650D/d = 1950(D/d)  nm

 

(b)

 

Let the nth bright fringe due to wavelength λ2 and (n – 1)th bright fringe due to wavelength λ1coincide on the screen. We can equate the conditions for bright fringes as:

 

nλ2 = (n-1) λ1

520n = 650n - 650

n = 5

Hence, the least distance from the central maximum can be obtained by the relation:

 

x = λ2D/d

   = 5 x 520 x D/d

   = 260 D/d  nm

Answer 2

Explanation:

I hope it Will help you..