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s344n2d4d5 [400]
3 years ago
8

Light of wavelength 600 nm passes though two slits separated by 0.25 mm and is observed on a screen 1.4 m behind the slits. The

location of the central maximum is marked on the screen and labeled y = 0.
At what distance, on either side of y = 0, are the m = 1 bright fringes ?
Physics
2 answers:
Arada [10]3 years ago
6 0

Answer:

y = 3.36 mm

Explanation:

The wavelength of the light, \lambda = 600 nm= 600 * 10^{-9} m

Separation between the two slits, d = 0.25 mm

d = 0.25 * 10^{-3} m

The separation between the light image and the screen, L = 1.4 m

m = 1

the distance of the bright fringes from the central maximum, y = \frac{m \lambda L}{d}

y = \frac{1 * 600 * 10^{-9} }{0.25 * 10^{-3} }

y = 0.00336 m

y = 3.36 mm

Anarel [89]3 years ago
5 0

Answer:

y(m=+1,-1)=3.36mm

Explanation:

We have to take into account the expression for the position of the fringes

y=\frac{m\lambda D}{d}

Where lambda is the wavelength of the light, D is the distance to the screen, m is the order of the fringe and d is the distance between slits.

By replacing we have

y=\frac{(1)(600*10^{-9}m)(1.4m)}{0.25*10^{-3}m}=3.36*10^{-3}m

There is a distance of 3.36mm to the secon maximum in the screen.

HOPE THIS HELPS!!

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