Question

In ydse, two slits are made 5 mm apart and the screen is placed 2 m away. What is the fringe separation when light of wavelength 500 nm is used?

Answer 1

Explanation:

xd /D = lambda

x = fringe separation

d = 5 × 10^-3 m

D = 2 m

lamda = 5 ×10^-7 m

Answer 2

Answer:

The answer is $5 \times 10^{-4} \mathrm{~m}$

Explanation:

The distance between any two consecutive bright fringes or two consecutive dark fringes exists named fringe spacing. Fringe spacing or thickness of a dark fringe or a bright fringe stands equivalent. It is indicated by Dx.

It is given that,  Two slits exist created 5 mm apart and the screen is placed 2 m away.

To find the fringe separation when the light of wavelength 500 nm is used.

Fringe width is given by $\beta=\frac{\lambda D}{d}$

$=\frac{500 \times 10^{-9} \times 1}{10^{-3}}$

$=0.5 \mathrm{~mm}=0.5 \times 10^{-3} \mathrm{~m}$

$=5 \times 10^{-4} \mathrm{~m}$

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