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Verizon [17]
3 years ago
12

Need help in the middle one

Physics
1 answer:
kondor19780726 [428]3 years ago
7 0

Answer:

Guysi hate math answer this guy plsss ssss

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A diffraction pattern is formed on a screen 130 cm away from a 0.420-mm-wide slit. Monochromatic 546.1-nm light is used. Calcula
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Answer:

The fractional Intensity \frac{I}{I_{max}} = 0.0146

Given:

wavelength of the light, \lambda = 546.1 nm = 546.1\times 10^{-9} m

slit and screen separation difference, D = 130 cm = 1.3 m

distance of the point from the center of the principal maximum, y = 4.10 mm = 0.041 m

slit width, d = 0.420 mm = 0.420\times 10^{-3}

Solution:

To calculate the fractional intensity, we use the given formula:

\frac{I}{I_{max}} = \frac{sin^{2}\delta }{\frac({\delta}{2})^{2}}             (1)

\delta = \frac{\pi }{\lambda}dsin\theta    

For very small angle:                                        

\delta = \frac{\pi dy}{\lambda D}                                  (2)

where

\delta = total phase angle

\theta = angle of deviation

Using eqn (2):

\delta = \frac{\pi \times 0.42\times 10^{-3}\times 4.1\times 10^{-3} }{546.1\times 10^{-9}\times 1.3} = 7.6202 radians

Now, using eqn (1):

\frac{I}{I_{max}} = \frac{sin^{2}(7.6202) }{(\frac{7.6202}{2})^{2}} = 0.0146

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