Test: Wave Optics - 2 - CUET Humanities MCQ

When the visible light from below the Sun's surface passes through the layers above it (the photosphere and the chromospheres), some of the light at particular wavelengths is absorbed by atoms and ions, so the missing lines are called Fraunhofer lines. At the time of the solar eclipse, there is no light from the central part (photosphere) of the sun on the earth. The light on the earth comes from chromospheres, which contains various elements of the excited gaseous state. The spectrum consists of bright emission lines against dark background. These bright lines correspond to the Fraunhofer lines observed in normal solar spectrum.

Number of fringes = (width of region/fringes width) or n = L/β.
Now, fringe width β is proportional to wavelength. Hence the new number of fringes will be   which is choice (2).

It is found that the width of central maxima is twice as that of a secondary maxima. Also, the intensity of the secondary maxima goes on decreasing with the order of maxima.

We know that the intensity is proportional to square of amplitude.

Fringe width, w = x_n+1 – x_n = Dλ/d
Since D is halved, w (fringe width) will become double.

The path difference is given by δ = dsin θ​​​​​​​. When L >> y, θ​​​​​​​ is small and we can make the approximation sinθ​​​​​​​ ≈ tanθ = y/L. Thus,

When one slit is covered then I₂ = 0.
∴ I'₀ = I
= I₀/4

I = I₀ cos^2θ
When θ lies between 0 and 2π^_, extinction occurs at π/2 and 3π/2^, and brightness occurs at 0 and π.

Since, the frequency n of the light does not change as light travels from air into glass, we have 
Therefore, 

Energy is neither destroyed nor created.
In case of interference, the superposition of two or more waves takes place resulting in a new wave pattern. So, energy is distributed in case of interference.