Monochromatic light of wavelength λ=620nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is observed on a screen 3.00 m from the slit.Part AIn terms of the intensity I0 at the peak of the central maximum, what is the intensity of the light at the screen at the distance 1.00 mm from the center of the central maximum?

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21-04-2020
Physics

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Monochromatic light of wavelength λ=620nm from a distant source passes through a slit 0.450 mm wide. The diffraction pattern is observed on a screen 3.00 m from the slit.Part AIn terms of the intensity I0 at the peak of the central maximum, what is the intensity of the light at the screen at the distance 1.00 mm from the center of the central maximum?

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-04-23T06:06:14+02:00

Answer:

The intensity of light from the 1mm from the central maximu is [tex]I = 0.822I_o[/tex]

Explanation:

From the question we are told that

The wavelength is [tex]\lambda = 620 nm = 620 *10^{-9}m[/tex]

The width of the slit is [tex]w = 0.450mm = \frac{0.45}{1000} = 0.45*10^{-3} m[/tex]

The distance from the screen is [tex]D = 3.00m[/tex]

The intensity at the central maximum is [tex]I_o[/tex]

The distance from the central maximum is [tex]d_1 = 1.00mm = \frac{1}{1000} = 1.0*10^{-3}m[/tex]

Let z be the the distance of a point with intensity I from central maximum

Then we can represent this intensity as

[tex]I = I_o [\frac{sin [\frac{\pi * w * sin (\theta )}{\lambda} ]}{\frac{\pi * w * sin (\theta )}{\lambda } } ]^2[/tex]

Now the relationship between D and z can be represented using the SOHCAHTOA rule i.e

[tex]sin \theta = \frac{z}{D}[/tex]

if the angle between the the light at z and the central maximum is small

Then [tex]sin \theta = \theta[/tex]

Which implies that

[tex]\theta = \frac{z}{D}[/tex]

substituting this into the equation for the intensity

[tex]I = I_o [\frac{sin [\frac{\pi w}{\lambda} \cdot \frac{z}{D} ]}{\frac{\pi w z}{\lambda D\frac{x}{y} } } ][/tex]

given that [tex]z =1mm = 1*10^{-3}m[/tex]

We have that

[tex]I = I_o [\frac{sin[\frac{3.142 * 0.45*10^{-3}}{(620 *10^{-9})} \cdot \frac{1*10^{-3}}{3} ]}{\frac{3.142 * 0.45*10^{-3}*1*10^{-3} }{620*10^{-9} *3} } ]^2[/tex]

[tex]=I_o [\frac{sin(0.760)}{0.760}] ^2[/tex]

[tex]I = 0.822I_o[/tex]

dsdrajlin dsdrajlin · Answer 2 · 2021-12-17T13:52:21+01:00

Monochromatic light of wavelength 620nm from a distant source passes through a slit 0.450 mm wide and is projected on a screen 3.00 m from the slit. The intensity of the light is 0.822 of the intensity at the peak of the central maximum.

What is light intensity?

Light intensity (I) refers to the strength or amount of light produced by a specific lamp source. It is the measure of the wavelength-weighted power emitted by a light source.

Monochromatic light of wavelength λ = 620nm from a distant source passes through a slit 0.450 mm wide (d). The screen is 3.00 m (D) from the slit.

Let x be the distance of a point with intensity I from the central maximum. We can relate it to the intensity at the central maximum (I₀) using the following expression.

[tex]I = I_0 [\frac{sin\frac{(\pi d)sin(\theta)}{\lambda } }{\frac{(\pi d)sin(\theta)}{\lambda }} ]^{2}[/tex] [1]

We can relate x and D in terms of sinθ.

sinθ = x/D [2]

For a very small angle, we can write sinθ as θ.

θ = x/D [3]

When the distance from the central maximum is x = 1.00 mm, substituting [3] in [1], we get:

[tex]I = I_0 [\frac{sin\frac{(\pi d)x}{\lambda D} }{\frac{(\pi d)x}{\lambda D }} ]^{2}\\\\I = I_0 [\frac{sin\frac{\pi (0.450 \times 10^{-3}m )(1.00\times 10^{-3}m )}{(620 \times 10^{-9}m ) (3.00m)} }{\frac{(\pi (0.450 \times 10^{-3}m)(1.00\times 10^{-3}m )}{(620 \times 10^{-9}m) (3.00m) }} ]^{2}\\\\I = 0.822 I_0[/tex]

Monochromatic light of wavelength 620nm from a distant source passes through a slit 0.450 mm wide and is projected on a screen 3.00 m from the slit. The intensity of the light is 0.822 of the intensity at the peak of the central maximum.

Learn more about light intensity here: https://brainly.com/question/3917542