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puteri [66]
2 years ago
12

Which is the destructive interference formula for diffraction grating problems? dsin = dcos = dcos = n dsin = n.

Physics
1 answer:
shusha [124]2 years ago
5 0

The destructive interference formula for diffraction grating problems is d sin \theta = (n+\frac{1}{2})\lambda.

<h3>What is the definition of destructive interference?</h3>

Destructive interference happens when the maxima of two waves are 180° out of phase a positive displacement of one wave is canceled exactly by a negative displacement of the other wave.

The formula for brighter patches resulting from constructive interference and darker patches resulting from destructive interference in a diffraction grating is:

\rm d sin \theta = n \lambda

The grating spacing is denoted by d, the angle of light is denoted by a the fringe order is denoted by n, and the wavelength is denoted by \rm \lambda.

The destructive interference formula is now based on the fact that destructive interference occurs between the fringes.

Hence the destructive interference formula for diffraction grating problems is d sin \theta = (n+\frac{1}{2})\lambda.

To learn more about destructive interference refer to the link;

brainly.com/question/16098226

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Two copper rods are separated by a small gap at B. Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the f
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Answer:

A)3.8196 * 10^{9} Newtons        B) 2.153 * 10^{9} Newtons

Explanation:

Force = Pressure * Area.

Pressure is given as  120GPa

Area = Cross Sectional Area of Rod = Area Of A circle = πR^{2}.

Area Expansivity β =    \frac{Change in Area}{Original Area * Temperature Rise}

Area Expansivity β = 2α

α =  16.9x10-6/ ° C.

Radius of Rod AB = Half 0f Diameter, 200mm = \frac{0.2m}{2} = 0.1 m

Radius of Rod BC = Half Of Diameter, 150mm = \frac{0.15}{2} = 0.075 m.

Note: To convert from millimeter to meter you divide by 1000.

Area of Rod AB = π * 0.1^{2} =  0.0314m^{2}

Area of Rod BC = π * 0.075^{2} = 0.0177m^{2}.

Change in Area = 2α * Original Area * Temperature Rise. Therefore for

Rod AB = 2 * 16.9*10^{-6} * 0.0314 * 30 = 3.183 * 10^{-5}m^{2}.

For Rod BC we have = 2 * 16.9*10^{-6} * 0.0177 * 30 = 1.794∈-5m^{2}.

The forces on each rods will be given by.

Force AB = Pressure * Area  of Rod AB

                 = 120GPa * 3.183 * 10^{-5} gives 3.8196 * 10^{9} Newtons.

Force BC = Pressure * Area of Rod BC

                = 120GPa * 1.794 * 10^{-5} gives 2.153 * 10^{9} Newtons

7 0
3 years ago
For a particular reaction, the change in enthalpy is 51kJmole and the activation energy is 109kJmole. Which of the following cou
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Answer

given,

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