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anyanavicka [17]
3 years ago
8

Coherent monochromatic light of wavelength l passes through a narrow slit of width a, and a diffraction pattern is observed on a

screen that is a distance x from the slit. On the screen, the width w of the central diffraction maximum is twice the distance x. What is the ratio a>l of the width of the slit to the wavelength of the light
Physics
1 answer:
geniusboy [140]3 years ago
3 0

Answer:

λ = a

Explanation:

This is a diffraction exercise that is described by the expression

          a sin θ = m λ

         sin θ  = m λ/ a

the first zero of the diffraction occurs for m = 1

        sin θ  = λ / a

 

angles are generally very small and are measured in radians

         sin θ  = θ  = y / x

we substitute

         \frac{y}{x}  = \frac{\lambda}{a}

the width of the central maximum is twice the distance to zero

         w = 2y

in the exercise indicate that this width is equal to twice the distance to the screen (2x)

          W = 2x

           2y = 2x

we substitute

          1 = λ/ a

          λ = a

we see that the width of the slit is equal to the wavelength used.

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Fnet = 0

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  • This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

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  • In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

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Light shined through a single slit will produce a diffraction pattern. Green light (565 nm) is shined on a slit with width 0.210
kondor19780726 [428]

Answer:(a)9.685 mm

(b)4.184 mm

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Given

Wavelength of light (\lambda )=565nm \approx 565\times 10^{-9}m

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=\frac{2\lambda L}{b}

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(b)Width of the first order bright fringe

Y_1=\frac{\lambda \times L}{b}

Y_1=\frac{565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}

Y_1=4.84mm

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2 years ago
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