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Bingel [31]
3 years ago
12

Two narrow slits, illuminated by light consisting of two distinct wavelengths, produce two overlapping colored interference patt

erns on a distant screen. The center of the eighth bright fringe in one pattern coincides with the center of the third bright fringe in the other pattern. What is the ratio of the two wavelengths?
Physics
1 answer:
vivado [14]3 years ago
6 0

Answer:

The ration of the two wavelength is  \frac{\lambda_1}{\lambda_2}  =  \frac{8}{3}

Explanation:

Generally two slit constructive interference can be mathematically represented as

      \frac{y}{L}  =  \frac{m *  \lambda}{d}

Where  y is the distance between fringe

           d  is the distance between the two slit

           L is the distance between the slit and the wall

           m is the order of the fringe

given that  y , L  , d  are constant  we have that

     \frac{m }{\lambda }  =  constant

So  

    \frac{m_1 }{\lambda_1 }  =  \frac{m_2 }{\lambda_2 }

So     m_1 = 8

  and  m_2 =  3

=>     \frac{m_2}{m_1}  =  \frac{\lambda_1}{\lambda_2}

=>     \frac{8}{3}  =  \frac{\lambda_1}{\lambda_2}

So

     \frac{\lambda_1}{\lambda_2}  =  \frac{8}{3}

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Answer:

Part a)

V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}

v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

y_{cm} = \frac{mR}{m + M}

Explanation:

PART A)

As we know that there is no external force on the system of two masses in horizontal direction

So here the two masses will have its momentum conserved in horizontal direction

So we have

mv + MV = 0

Also we know that here no friction force on the system so total energy will always remains conserved

So we have

\frac{1}{2}mv^2 + \frac{1}{2}MV^2 = mgR

now we have

\frac{1}{2}m(\frac{MV}{m})^2 + \frac{1}{2}MV^2 = mgR

\frac{1}{2}MV^2(\frac{M}{m} + 1) = mgR

so we have

V = \sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}

and another block has speed

v = \frac{M}{m}\sqrt{\frac{2\frac{m}{M}gR}{(\frac{M}{m} + 1)}}

Part b)

Since on the block wedge system there is no external force in horizontal direction so the Center of mass will not move in horizontal direction but in vertical direction it will move

so displacement in Y direction is given as

y_{cm} = \frac{mR}{m + M}

7 0
3 years ago
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