Physics, asked by Firerage7660, 11 months ago

In ydse if width of slits arebin the ratio 4:9 the ratio of the intensity of madima to intensity at minima will be

Answers

Answered by namrata6969
0

Answer:

The ratio of the minimum and maximum intensity is \dfrac{1}{4}

4

1

.

Explanation:

Given that,

The slit widths are in the ratio 1 : 9,

We know that,

Width is directly proportional to the intensity

\dfrac{\beta_{1}}{\beta_{2}}=\dfrac{I_{1}}{I_{2}}

β

2

β

1

=

I

2

I

1

\dfrac{I_{1}}{I_{2}}=\dfrac{1}{9}

I

2

I

1

=

9

1

The intensity is directly proportional to square of the amplitude.

\dfrac{I_{1}}{I_{2}}=(\dfrac{A_{1}^2}{A_{2}^2})

I

2

I

1

=(

A

2

2

A

1

2

)

\dfrac{A_{1}}{A_{2}}=\dfrac{1}{3}

A

2

A

1

=

3

1

The maximum intensity is

I_{max}=(A_{1}+A_{2})^2I

max

=(A

1

+A

2

)

2

I_{max}=(1+3)^2I

max

=(1+3)

2

I_{max}=16I

max

=16

The minimum intensity is

I_{min}=(A_{1}-A_{2})^2I

min

=(A

1

−A

2

)

2

I_{min}=(1-3)^2I

min

=(1−3)

2

I_{min}=4I

min

=4

The ratio of the minimum and maximum intensity will be

\dfrac{I_{min}}{I_{max}}=\dfrac{A_{2}}{A_{1}}

I

max

I

min

=

A

1

A

2

\dfrac{I_{min}}{I_{max}}=\dfrac{4}{16}

I

max

I

min

=

16

4

\dfrac{I_{min}}{I_{max}}=\dfrac{1}{4}

I

max

I

min

=

4

1

Hence, The ratio of the minimum and maximum intensity is \dfrac{1}{4}

4

1

.

Explanation:

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