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xenn [34]
3 years ago
13

Determine the dimensions (a, b) of an air-filled rectangular waveguide that providessingle mode operation over the frequency ran

ge from 9 GHz to 14 GHz.
Physics
1 answer:
Dima020 [189]3 years ago
8 0

Answer: the required dimensions (a, b) are; ( a= 1.667 cm, b = 1.0714 cm )

Explanation:

Given the data in the question;

air-filled rectangular waveguide  

μ₀ = C

m = 1 --- { for single mode operation}

f₀ = c/za        {we have equation}

denote the f₀ = 9 GHz  

f₀  = Carrier frequency

f₁₀ = 14 Ghz        Range or cut off frequency

so we have to find;

a = c/2f₀ and b = c/2f₁₀

so

f₀ = 9 GHz

a = c/2f₀  

a = (3 × 10⁸) / ( 2× 9 × 10⁹)

a = 0.01666 cm

a = 1.667 cm

f₁₀ = 14 GHz

b = c/2f₀  

b = (3 × 10⁸) / ( 2× 14 × 10⁹)

b = 0.01071 m

b = 1.0714 cm

Therefore the required dimensions (a, b) are; ( a= 1.667 cm, b = 1.0714 cm )

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Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (As
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Answer:

a

   \lambda = 1.667 nm

b

     \theta  =  0.8681^o

Explanation:

From the question we are told that

   The distance of separation is d  =  0.220 \ mm  =  0.00022 \ m

    The  is distance of the screen from the slit is  D   =  2.60 \ m

    The distance between the central bright fringe and either of the adjacent bright   y  =  1.97 cm  =  1.97 *10^{-2}\ m

Generally  the condition for constructive interference is  

      d sin \tha(\theta ) =  n \lambda

From the question we are told that small-angle approximation is valid here.

So    sin (\theta ) = \theta

=>        d \theta  =  n \lambda

=>        \theta =  \frac{n *  \lambda }{d }

Here n is the order of maxima and the value is  n =  1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright  is mathematically represented as

         y  =  D * sin (\theta )

From the question we are told that small-angle approximation is valid here.

So

       y  =  D * \theta

=>   \theta  =  \frac{ y}{D}

So

     \frac{n *  \lambda }{d } = \frac{y}{D}

     \lambda =\frac{d * y }{n * D}

substituting values

       \lambda =  \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }

        \lambda = 1.667 *10^{-6}

        \lambda = 1.667 nm

In the b part of the question we are considering the next set of bright fringe so  n=  2

    Hence

     dsin (\theta ) =  n \lambda

    \theta  =  sin^{-1}[\frac{ n  *  \lambda }{d} ]

    \theta  =  sin^{-1}[\frac{ 2  *  1667 *10^{-9}}{ 0.00022} ]

    \theta  =  0.8681^o

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