Answer to Question #206297 in Mechanics | Relativity for KABIR

Question

• In Young's double - slit experiment, a monochromatic parallel light of wavelength, λ=550 nm illuminates the double slits with separation of a=2×10⁻⁴m. The distance between the viewing screen and the barrier containing the double slits is D=2m. Find : (1) The distance between the pair of the tenth order bright fringes that are on both sides of the central maximum? (2) If one slit is covered by a glass sheet of thickness =6.6×10⁻⁵ m and index of refraction n=1.58, the zeroth maximum will move to a new position which corresponds to the original Kth order bright fringe. What is the value of K?

Answer 1

Gives

D=2 m

$\mu$ =1.58

$\lambda=500nm$

$d=2\times10^{-4} m$

(n)=10

$x=\frac{n\lambda D}{d}$

$x=\frac{10\times 550\times10^{-9} \times2}{2\times10^{-4}}$

x=0.055m

Part(b)

$t=6.6\times10^{-5}m$

$2(\mu -1)t=2x$

$$x=(1.58-1)\times6.6\times10^{-5}=3.828\times10^{-5}m$$

Find k Value

$x=\frac{k\lambda D}{d}$

Put value

$k=\frac{xd}{\lambda D}$

$k=\frac{3.828\times10^{-5} \times2\times10^{-4}}{500\times10^{-9}\times2}=8$