Physics: You pass 633 nm laser light through a narrow single slit of width 0.24 mm and observe the diffraction pattern on a screen 6.0 m away. If the intensity at the central bright is Io , what is the intensity at a point on the screen 3.0 mm from the center of the pattern?

You pass 633 nm laser light through a narrow single

kim643

10.02.2022

The intensity is I = 0.0175I_o

Explanation:

From the question we are told that

The wavelength is $\lambda = 633 \ nm = 633*10^{-9} \ m$

The width of the slit is $d = 0.24 \ mm = 0.00024 \ m$

The distance from the screen is $D = 6.0 \ m$

The distance of the position considered from the center is $y = 3.0 \ mm = 0.003 \ m$

Generally the intensity from at a point on the screen 3.0 mm from the center of the pattern is

$I = I_o * \frac{sin^2 [\frac{\pi d * y}{ \lambda D } ]}{[\frac{\pi d * y }{\lambda D} ]^2}$

Here I_o is the intensity of the central bright fringe

=> $I = I_o * \frac{sin^2 [\frac{3.142 * 0.00024 * 0.003}{ 633*10^{-9} * 6 } ]}{[\frac{3.142*0.00024 * 0.003 }{633*10^{-9} * 6} ]^2}$

=> I = 0.0175I_o

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brittanyjacob8

11.04.2023

The anvil will travel at the speed of 44.27 m/s

Explanation:

Given:

mass of anvil, m = 20 kg

height, h = 100 m

Here , work done is equal to change in kinetic energy. Here initial kinetic energy is zero. Hence work done is equal to final kinetic energy.

Work done = force * distance

= mg * h .............. (equation 1)

(here work done vertically is considered hence a= g = 9.8 m/ $s^{2}$ and distance = h)

Also ,

work done = final kinetic energy

= $\frac{1}{2}$ $mv^{2}$ ...............(equation 2)

from equation 1 and 2 , we get

mgh = $\frac{1}{2}$ $mv^{2}$

$v^{2}$ = 2gh

= 2 * 9.8 * 100

= 1960

v = 44.27 m/s

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