# You illuminate a slit with a width of 77.7 μm with a light of wavelength 721 nm and observe the resulting diffraction pattern on a screen th

You illuminate a slit with a width of 77.7 μm with a light of wavelength 721 nm and observe the resulting diffraction pattern on a screen that is situated 2.83 m from the slit. What is the width, in centimeters, of the pattern’s central maximum

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1. haidang

The width is  $$Z = 0.0424 \ m$$

Explanation:

From the question we are told that

The width of the slit is $$d = 77.7 \mu m = 77.7 *10^{-6} \ m$$

The wavelength of the light is  $$\lambda = 721 \ nm$$

The position of the screen is  $$D = 2.83 \ m$$

Generally angle at which the first minimum  of the interference pattern the  light occurs  is mathematically  represented as

$$\theta = sin ^{-1}[\frac{m \lambda}{d} ]$$

Where m which is the order of the interference is 1

substituting values

$$\theta = sin ^{-1}[\frac{1 *721*10^{-9}}{ 77.7*10^{-6}} ]$$

$$\theta = 0.5317 ^o$$

Now the width of first minimum  of the interference pattern is mathematically evaluated as

$$Y = D sin \theta$$

substituting values

$$Y = 2.283 * sin (0.5317)$$

$$Y = 0.02 12 \ m$$

Now the width of  the  pattern’s central maximum is mathematically evaluated as

$$Z = 2 * Y$$

substituting values

$$Z = 2 * 0.0212$$

$$Z = 0.0424 \ m$$