OneTime: Interference NAT Level - 2 - Physics MCQ

λ_red = 632.82 nm

λ_red  λ_green → (D) and (E) are FALSE.
Constructive path difference = Nλ

N_gλ_g = N_rλ_r

λ_g = N_rλ_r / N_g

= (85,865 × 632.82) / 100,000

≈ (8.6 × 10⁴ × 6.3 × 10²) / 10⁵

= 541.8 

= 542 (approx)

The single slit interference equation for bright fringes is given by  (for small angles) where d is the width of the slit & in is an integer.
Since c = λv  one can relate the above equation to get 
Increasing frequency would decrease the angle thus, the fringes would get closer together. Increasing the frequency by a factor of 2 would decrease the separation by 2.
The correct answer is: 2

ΔL = 5 cm = 5 × 10⁻² m
ΔN = 40 fringes
λ_vac = 500 nm  = 5 × 10⁻⁷ m

40 = [2(5 × 10⁻²)/(5 × 10⁻⁷)](n − 1)
40 = (2 × 10⁵)(n − 1)
40/(2 × 10⁵) = n − 1
2 × 10⁻⁴ = n − 1
n = 0.0002 + 1 = 1.0002
 

Assuming the boundaries are vacuum-oil, oil-glass, then there will be a phase shift at the vacuum oil boundary since the refractive index of vacuum n_vac < n_oil  & another phase shift at the oil glass boundary because n_oil n_glass That tells us that constructive interference happens for integer wavelength mλ.
Light crosses a distance d in  but since its speed is lower in oil,

Then

∴ The condition for constructive interference becomes 2 n_oild = mλ
The next minimum thickness other than zero is when m = -1
So, 
The correct answer is: 200

We need a minimum of the single slit diffraction to coincide with a maximum of the double slit interference pattern (which result in a missing interference maximum)

[angulars location of minima of single slit diffraction]

[Angular locations of the maxima of double slit interference]
Ratio of 2 equations gives

So, minimum integer ratio is 2/1

The correct answer is: 2

In order for the thin film layer to be non reflecting, it must cancel the reflected wavelength as in a destructive interference the change in wavelength as

Since, the wave changes phase by π at the interface below air and the coating, and again at the second interface below coating and glass
(Assume n_air < n_coating < n_glass). Destructive interference is thus given by a half integer wavelength change mλ/2 where the smallest change is 

The correct answer is: 0.25

For a thin film of thickness t, one can easily find the condition for interference phenomenon. Since the light has to travel approximately 2t to get back to the original incidence interface, one has 2t = mλ. However, since the light changes phase at the interface between air & glass, the condition for constructive interference becomes,

Minimum thickness 
The correct answer is: 122

Minimum intensity Imin​=(√I₁ ​− √I₂)²
(√I ​− √4I​)²
= I

The key equation involved is 
For small angles,  
This blur is due to both the size of the hole and the diffraction, the arc length of the diffraction is approximately
where the factor of 2 comes from the fact that the arc length angle is symmetrical about the diffraction axis, thus twice the diffraction angle.
Define Blur equation taking the first derivative, w.r.t.
to d

The correct answer is: 1.414

The ratio of intensities of light proceeding from the slits, I₁: I₂ = 4 : 1

Therefore, the ratio of amplitudes of light proceeding from the slits,

a₁:a₂ = √I₁: √I₂ = 2:1

The amplitudes at the maximum and minimum of the interference pattern are therefore in the ratio (a₁+a₂) : (a₁–a₂) = 3:1

The intensities at the maximum and the minimum are in the ratio3²:1² = 9:1