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This mock test of Inteference NAT Level - 2 for Physics helps you for every Physics entrance exam.
This contains 10 Multiple Choice Questions for Physics Inteference NAT Level - 2 (mcq) to study with solutions a complete question bank.
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*Answer can only contain numeric values

QUESTION: 1

A Michelson interferometer is configured as a wave meter as shown in the figure below may be used to compare the wavelengths of 2 laser with high precision when the mirror is the right arm of the interferometer is translated through a distance d, 1,00,000 interference fringes pass across the detector for green light and 85,865 fringes pass across the detector for red (λ = 632.82 nm) light. What is the wavelength of the green laser light in nm?

Solution:

*λ*_{red}* *= 632.82 nm

*λ*_{red } *λ*_{green }→ (D) and (E) are FALSE.

Constructive path difference = *N**λ*

*N*_{g}*λ*_{g}_{ }= *N*_{r}*λ*_{r}

*λ*_{g}_{ }= *N*_{r}*λ*_{r }/ *N*_{g}

= (85,865 × 632.82) / 100,000

≈ (8.6 × 10^{4 }× 6.3 × 10^{2}) / 10^{5}

= 541.8

= 542 (approx)

*Answer can only contain numeric values

QUESTION: 2

Light from a laser falls on a pair of very narrow slits separated by 0.5 μm & bright fringes separated by 1.0 mm are observed on a distant screen if the frequency of the laser light is doubled, how much times the separation of the bright fringes decrease(answer in integer) ?

Solution:

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

*Answer can only contain numeric values

QUESTION: 3

A gas filled cell of length 0.5 cm is inserted in one arm of the michelson interferometer as shown in the figure below. The interferometer is illuminated by light of wavelength 500 nm. As the gas is evaluated from the cell, 40 fringes cross a point in the field of view. Find the refractive index of this gas

Solution:

Δ*L *= 5 cm = 5 × 10^{−2 }m

Δ*N** *= 40 fringes

*λ*_{vac}* *= 500 nm = 5 × 10^{−7 }m

40 = [2(5 × 10^{−2})/(5 × 10^{−7})](*n* − 1)

40 = (2 × 10^{5})(*n* − 1)

40/(2 × 10^{5}) = *n* − 1

2 × 10^{−4}^{ }= *n* − 1

*n*^{ }= 0.0002 + 1 = 1.0002

*Answer can only contain numeric values

QUESTION: 4

Blue light of wavelength 480 nm is most strongly reflected from a thin film of oil on a glass slide when viewed near normal incidence. Assuming that the index of refraction of the oil is 1.2 & that of glass is 1.6, what is the minimum thickness of the oil film in nm (other than zero)?

Solution:

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_{oil}d = mλ

The next minimum thickness other than zero is when m = -1

So,

The correct answer is: 200

*Answer can only contain numeric values

QUESTION: 5

In a double slit interference experiment d is the distance between the centres of the slits and w is the width of each slit. Find the minimum integer ratio d/ω for which an interference maximum of incident plane waves on a distant screen will be missing.

Solution:

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

*Answer can only contain numeric values

QUESTION: 6

It is necessary to coat a glass lens with a non reflecting layer if the wavelength of the light in the coating is λ, the best choice is a layer of material having an index of refraction between those of glass & air and a thickness of what fraction of λ ?

Solution:

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

*Answer can only contain numeric values

QUESTION: 7

Consider two horizontal glass plates with a thin film of air between them. Find the minimum value of the thickness of the film in nm of air for which the film as seen by the reflected light will appear bright if it is illuminated normally from above by blue light of wavelength 485 nm?

Solution:

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

*Answer can only contain numeric values

QUESTION: 8

Two coherent monochromatic light beams of intensities l and 4l are superposed. What is the minimum possible intensities in the resulting beam?

Solution:

Minimum intensity Imin=(√I_{1 }− √I_{2})^{2}

(√I − √4I)^{2}

= I

*Answer can only contain numeric values

QUESTION: 9

The screen of a pinhole camera is at a distance D from the pinhole. Which has a diameter d the light has an effective wavelength For what value of d in terms of will the image be sharpest

Solution:

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

*Answer can only contain numeric values

QUESTION: 10

A student prepared a young’s double slit by drawing two parallel lines (with a separation less than a millimeter) on a smoked glass plate. Accidently, he used two different pins to draw the lines so that the widths of the slits (regions from where smoke was removed by the pin) were in the ratio 4:1. What will be the ratio of the interference maximum to that at the interference minimum?

Solution:

The ratio of intensities of light proceeding from the slits, I_{1}: I_{2} = 4 : 1

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

a_{1}:a_{2} = √I_{1}: √I_{2} = 2:1

The amplitudes at the maximum and minimum of the interference pattern are therefore in the ratio (a_{1}+a_{2}) : (a_{1}–a_{2}) = 3:1

The intensities at the maximum and the minimum are in the ratio3^{2}:1^{2} = **9:1**

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