Answer:
The wavelengths of these two fringes is 589.61 x 10∧9
Step-by-step explanation:
d = slit width = 0.001/1000 = 1 x 10-6 m
X1 = Position of first fringe = 0.7288 m
X2 = Position of second fringe = 0.73 m
D = distance of the screen
tan\theta1 = X1/D \Rightarrow\theta1 = tan-1(X1/D ) = tan-1(0.7288/1 ) = 36.085
tan\theta2 = X2/D \Rightarrow\theta2 = tan-1(X2/D ) = tan-1(0.73/1 ) = 36.129
position of first bright fringe is given as
d Sin\theta1 = n\lambda1
for bright fringe , n = 1
d Sin\theta1 = \lambda1
\lambda1 = (1 x 10-6 ) Sin36.085 = 589 x 10-9
position of second bright fringe is given as
d Sin\theta2 = n\lambda2
for bright fringe , n = 1
d Sin\theta2 = \lambda2
\lambda2 = (1 x 10-6 ) Sin36.129 = 589.61 x 10-9