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3 years ago

Diffraction of white light with a single slit produces bright lines of different colors.What is the color of the central image?

Physics

1 answer:

olya-2409 [2.1K]3 years ago

7 0

Answer:

white

Explanation:

This experiment causes lines of different colors because the constructive interference depends on the wavelength, so for different wavelength, the position of the maximum will be different.

Now, in diffraction fo a single slit we always have a central maximum (for all the colors)

This means that after the diffraction, we will have that all the colors have a maximum in the center, which will produce white light again, then the color of the central image is white

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A, D

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2 years ago

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An aluminum block of mass 12.00 kg is heated from 20 c to 118 c. if the specific heat of aluminum is 913 j-1 kg k-1 then how muc

Jobisdone [24]

Answer:

107 kJ

Explanation:

12.00 kg * 913 J /kg-C * (118-20)°C ( I corrected spec heat units)

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1 year ago

A tank has a gate that automatically opens if the water levelhis high enough. The gate has a squarecross section of side1m and c

umka2103 [35]

Answer:

The gate will open if the height of water is equal to or more than 0.337m.

Explanation:

From the diagram attached, (as seen from the reference question found on google)

The forces are given as

Force on OA

$F_1=P A_1\\F_1=\rho g \bar{h} A_{OA}$

Here

ρ is the density of water.
g is the gravitational acceleration constant
$\bar{h}$ is the equivalent height given as

$\bar{h}=h+\frac{0.6}{2}\\\bar{h}=h+0.3$

$A_{OA}$ is the area of the OA part of the door which is calculated as follows:

$A_{OA}=L\times W\\A_{OA}=1\times 0.6\\A_{OA}=0.6 m^2$

The Force is given as

$F_1=0.6\rho g[h+0.3]$

Force on OB

$F_2=P A_2\\F_2=\rho g \bar{h} A_{OB}$

Here

ρ is the density of water.
g is the gravitational acceleration constant
$\bar{h}$ is the equivalent height given as

$\bar{h}=h+0.6+\frac{0.4}{2}\\\bar{h}=h+0.8$

$A_{OA}$ is the area of the OB part of the door which is calculated as follows:

$A_{OB}=L\times W\\A_{OB}=1\times 0.4\\A_{OB}=0.4 m^2$

The Force is given as

$F_2=0.4\rho g[h+0.8]$

Now the moment arms are given as

$\bar{y}_a=\bar{h}+\frac{I}{A\bar{h}}\\\bar{y}_a=h+0.3+\frac{\frac{1}{12}\times 0.6^3 \times 1}{0.6 \times[h+0.3]}\\\bar{y}_a=h+0.3+\frac{0.03}{h+0.3}$

$\bar{y}_b=\bar{h}+\frac{I}{A\bar{h}}\\\bar{y}_b=h+0.8+\frac{\frac{1}{12}\times 0.4^3 \times 1}{0.4 \times[h+0.8]}\\\bar{y}_b=h+0.8+\frac{0.0133}{h+0.8}$

Taking moment about the point O as zero

$F_1(h+0.6-\bar{y}_a)=F_2(\bar{y}_b-h+0.6)\\F_1(h+0.6-h-0.3-\frac{0.03}{h+0.3})=F_2(h+0.8+\frac{0.0133}{h+0.8}-h-0.6)\\F_1(0.3-\frac{0.03}{h+0.3})=F_2(0.2+\frac{0.0133}{h+0.8})\\0.6\rho g[h+0.3](0.3-\frac{0.03}{h+0.3})=0.4\rho g[h+0.8](0.2+\frac{0.0133}{h+0.8})\\0.6[h+0.3](0.3-\frac{0.03}{h+0.3})=0.4[h+0.8](0.2+\frac{0.0133}{h+0.8})\\0.18h -0.054-0.018=0.08h+0.064+0.00533\\h=0.337 m$

So the gate will open if the height of water is equal to or more than 0.337m.

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3 years ago

What is the relationship between the angle of incidence and the angle of refraction when light passes from a medium where its sp

Bogdan [553]

Answer:

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