Question

What is the width of the interference fringes?

• A. 1.5 x 10${}^{-3}$ m
• B. 3 x 10${}^{-3}$ m
• C. 2 x 10${}^{-4}$ m
• D. 2 x 10${}^{-2}$ m

Answer 1

Answer: (A)

1.5 x 10${}^{-3}$ m

We have,
$d=2(\mu -1)Al$
Given $\mu =1.5$, $A=2\times {10}^{-3}$ and $l=1$
$⟹d=2(1.5-1)x2x{10}^{-3}x1=2x{10}^{-3}m$
Also $D=4+1=5m$
We also have $\beta =\frac{\lambda D}{d}$
$⟹\beta =\frac{6000\times {10}^{-10}\times 5}{2\times {10}^{-3}}=1.5\times {10}^{-3}m$