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3 years ago

A zebra starts from rest and accelerates at 1.9 m/s how far has a zebra gone in five seconds

Physics

1 answer:

Maksim231197 [3]3 years ago

5 0

To that multiply 1.9 by 5 to get 9.5. The zebra has gone 9.5 m.

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Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’2) mm apart and then falls on a screen 7

IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

$sin \theta=\frac{n \lambda}{d}$

where

n is the order of the maximum

$\lambda$ is the wavelength

$a$ is the distance between the slits

In this problem,

n = 5

$\lambda=525 nm =5.25\cdot 10^{-7} m$

$a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m$

So we find

$\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}$

And given the distance of the screen from the slits,

$D=75.0 cm = 0.75 m$

The distance of the 5th bright fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 3.62^{\circ}=0.047 m = 4.7 cm$

B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

$sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}$

To find the angle corresponding to the 8th dark fringe, we substitute n=8:

$\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}$

And the distance of the 8th dark fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm$

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