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Deffense [45]
3 years ago
11

(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W

hat is its minimum width if it produces 50 minima? (c) 1000 minima?
Physics
1 answer:
Kitty [74]3 years ago
4 0

Answer:

The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

D\sin\theta=n\lambda

D=\dfrac{n\lambda}{\sin\theta}

Where, D = width of slit

\lambda = wavelength

Put the value into the formula

D=\dfrac{n\lambda}{\sin\theta}

Here, \sin\theta should be maximum.

So. maximum value of \sin\theta is 1

Put the value into the formula

D=\dfrac{1\times\lambda}{1}

D=\lambda

(b). If the minimum number  is 50

Then, the width is

D=\dfrac{50\times\lambda}{1}

D=50\lambda

(c). If the minimum number  is 1000

Then, the width is

D=\dfrac{1000\times\lambda}{1}

D=1000\lambda

Hence, The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

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Answer:

T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N

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           sin 63 = T_{2y} / T₂

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   - T₁ₓ + T₂ₓ = 0

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   T1 = T2 cos 63 / cos 41                (1)

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      T_{1y} + T_{2y} - T3 = 0

       T₁ sin 41 + T₂ sin 63 = T₃          (2)

to solve the system we substitute equation 1 in 2

        T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W

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          T₂ = 200 / (cos 63 tan 41 + sin 63)

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            T₁ = T₂ cos 63 / cos 41

             T₁ = 155.6 cos63 / cos 41

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             T₂ = 155.6 N

             T₃ = 200 N

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