Question

asked Oct 13, 2021 142k views

2 Answers

David Sanford · Answer 1 · 2021-10-16T06:43:57+0000

Answer:

a

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

$\theta = 39.60^o$

b

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

c

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

Step-by-step 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

$\theta = sin^(-1) ((\lambda_3)/(d) )$

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

= sin^(-1) (0.6374)

$\theta = 39.60^o$

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

i.e
$f = (c)/(\lambda)$

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

$\lambda_(new) = (512 *10^(-9))/(0.945)$

$= 0.542 \mu m$

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

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

= 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

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

$\Delta z=0.3906 m$

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