The width of the central bright fringe <u>becomes wider</u> in the resulting diffraction pattern of a single-slit diffraction experiment.
<h3>What is diffracted light?</h3>
The act of bending light around corners such that it spreads out and illuminates regions where a shadow is anticipated is known as diffraction of light. In general, since both occur simultaneously, it is challenging to distinguish between diffraction and interference. The diffraction of light is what causes the silver lining we see in the sky. A silver lining appears in the sky when the sunlight penetrates or strikes the cloud.
<h3>What precisely is single slit diffractive?</h3>
The single-slit diffraction experiment allows us to examine the phenomena of light bending, or diffraction, which enables coherent light from a source to interfere with itself and generate the diffraction pattern, a recognizable pattern on the screen. When the sources are small enough to be relative to the wavelength of light, diffraction is seen.
Learn more about diffraction
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Answer:
4.99 mg of vitamin C are in the beaker.
Explanation:
Given that,
Weight of vitamin = 0.0499 g
Molar mass = 176.124 g/mol
Weight of water = 100.0 ml
We need to calculate the mg of vitamin C in the beaker
We dissolve 0.0499 g vitamin C in water to from 100.0 ml solution.
100 ml solution contain 49.9 mg vitamin C
Now, we take 10 ml of this vitamin C solution in breaker
Since, 100 ml solution =49.9 mg vitamin C
Therefore,
Hence, 4.99 mg of vitamin C are in the beaker.
Answer:
Density = 1.1839 kg/m³
Mass = 227.3088 kg
Specific Gravity = 0.00118746 kg/m³
Explanation:
Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³
Now, from tables, density of air at 25°C is 1.1839 kg/m³
Now formula for density is;
ρ = mass(m)/volume(v)
Plugging in the relevant values to give;
1.1839 = m/192
m = 227.3088 kg
Formula for specific gravity of air is;
S.G_air = density of air/density of water
From tables, density of water at 25°C is 997 kg/m³
S.G_air = 1.1839/997 = 0.00118746 kg/m³
When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted ray, and it is given by:
(2)
where
is the refractive index of the second medium and
is the refractive index of the first medium.
We can find the ratio
by using Snell's law:
(1)
where
is the angle of incidence
is the angle of refraction
By using the data of the problem and re-arranging (1), we find
and if we use eq.(2) we can now find the value of the critical angle:
According to cuneiform tablets in the ancient world, straight lines cannot cross, and no motion in the world is not relative. Btw...I KNOW!!! GOT MILK???