Physics, 04.04.2020 14:36 Emiann222

In a double-slit interference experiment, the slit separation is 2.41 μm, the light wavelength is 512 nm, and the separation between the slits and the screen is 4.45 m. (a) What is the angle between the center and the third side bright fringe? If we decrease the light frequency to 94.5% of its initial value, (b) does the third side bright fringe move along the screen toward or away from the pattern's center and (c) how far does it move?

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Answer from: brodybb5515

The angle between the center and the third side bright fringe is

$\theta = 39.60^o$

The third side bright fringe move away from the pattern's center

The distance by which it moves away is $\Delta z=0.3906 m$

Explanation:

From the question the

The wavelength is $\lambda = 512nm$

In the first question we a asked to obtain the angle between the center and the third side bright fringe

since we are considering the third side of the bright fringe the wavelength of light on the three sides would be evaluated as

$\lambda_{3} = 3 * 512nm$

The slit separation is given as $d = 2.41 \mu m$

The angle between the center and the third side bright fringe is

$\theta = sin^{-1} (\frac{\lambda_3}{d} )$

$\theta = sin^{-1} (\frac{3 *512*10^{-9}}{2.24*10^{-6}} )$

$= sin^{-1} (0.6374)$

$\theta = 39.60^o$

When the frequency of the light is reduced the wavelength is increased

i.e $f = \frac{c}{\lambda}$

and this increase would cause the third side bright to move away from the pattern's center

Now from the question frequency is reduce to 94.5% this mean that the wavelength would also increase by the same as mathematically represented below

$\lambda_{new} = \frac{512 *10^{-9}}{0.945}$

$= 0.542 \mu m$

The angle between the center and the third side bright fringe is for new wavelength

$\theta = sin^{-1} (\frac{3 *512*10^{-9}}{2.41*10^{-6}} )$

= 42.46^o

The distance traveled away from the pattern's center is mathematically represented as

$z = A tan \theta$

Where A is the separation between the slits and the screen

$\Delta z = 4.45(tan 42.46 - tan39.60 )$

$\Delta z=0.3906 m$

Answer from: Quest

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?

Answer from: Quest

v 154.9= 0.9 length length s little over 154.9cm

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In a double-slit interference experiment, the slit separation is 2.41 μm, the light wavelength is 51...