Question

solve this q........​

Answer 1

.•♫•heya pglu •♫•.

since settle maximum of both the interference pattern coincide therefore,

m1D lemda1 /d = m2D lemda2/d

m1 lemda1= m2 lemda 2

lemda2 = m1 lemda 2/m2

= 3×590/4 nm

= 442.5nm

Answer 2

Answer:

Given :

Wavelength (1) = lambda1 = 5.9 nm = 59 A°

fringe width (1) = X1 = 3

fringe width (2) = X2 = 4

Wavelength (2) = ?

Since, the interference pattern of lights coincides therefore,

X1*D*lambda1/d = X2*D*lambda2/d

X1*lambda1 = X2*lambda2

=> 3*59 = 4*lambda2

=> lambda2 = 3*59/4 = 177/4 = 44.25 A°

Thus, Wavelength (2) = 44.25 A°