Wave Optics

Relation between intensities is

Therefore, The correct answer is (C) :

The correct answer is (B) : 2

To solve for the difference in resultant intensities at points A and B, we begin by applying the formula for intensity in an interference pattern where two beams of light interfere: I_resultant = I₁ + I₂ + 2√(I₁I₂)cos(δ), where δ is the phase difference.

Given:

• Intensities: I₁ = I, I₂ = 4I
• Phase difference at point A, δ_A = π/2
• Phase difference at point B, δ_B = π/3

Step 1: Calculate I_resultant at point A:

• Using δ_A = π/2, cos(π/2) = 0
• I_resultant,A = I + 4I + 2√(I*(4I))*0 = 5I

Step 2: Calculate I_resultant at point B:

• Using δ_B = π/3, cos(π/3) = 1/2
• I_resultant,B = I + 4I + 2√(I*(4I))*(1/2)
• Calculate: I_resultant,B = 5I + 2√(4I²)*(1/2) = 5I + 2(2I)*(1/2) = 5I + 2I = 7I

Step 3: Find the difference between the intensities:

• Difference: |I_resultant,B - I_resultant,A| = |7I - 5I| = 2I

The value of x is 2, confirming it falls within the given range (2, 2).

The correct answer is 35

The correct answer is (C) : 3.13 × 10⁵ ms^–1

λ_obs = λ_source
For ν << C,
= 1 +
⇒ ν = × 3 × m/s
⇒ ν ≃ 3.13 × m/s

To find the angular divergence corresponding to the second order minima for a single-slit diffraction, we use the formula for minima in single-slit diffraction given by:
,
where is the slit width, is the angular position of the minima, is the order of the minima, and is the wavelength of the light.
For the second order minima, , the slit width , and the wavelength .
Substituting these values:


.
For small angles, , thus .
Therefore, the angular divergence for the second order minima is , which calculates to , comfortably within the expected range of when multiplied out to the problem's scale, confirming the solution's accuracy.

To solve this problem, we need to understand the concept of polaroid sheets and how light intensity varies when a polaroid sheet is rotated between two crossed polaroids.

When two Polaroids are crossed (90 degrees apart), ideally, no light should pass through. This is because the plane of polarization is perpendicular to each other, blocking any light.

Inserting a third polaroid sheet at an angle between the two crossed polaroids will allow some light to pass through due to the mechanism of polarization, which follows Malus's Law.

Malus's Law: The intensity of light transmitted through a Polaroid is given by:

where is the initial intensity of the light, and is the angle between the light's initial polarization direction and the axis of the Polaroid.

The problem asks for the angle at which the transmitted light intensity is maximum when a Polaroid sheet is rotated between two crossed Polaroids. The maximum transmitted intensity occurs when the intervening Polaroid is at an angle of to each of the crossed polaroids.

Let's apply this concept:

• between the first Polaroid and the middle sheet.
• The remaining rotation, , is between the middle sheet and the last Polaroid.

Therefore, for maximum transmitted light intensity, the middle Polaroid should be rotated by .

Hence, the correct answer is .

To find the width of each slit that results in 10 maxima of the double-slit pattern within the central maximum of the single-slit pattern, we begin by analyzing the interference and diffraction conditions.

For a double-slit, the condition for maxima is:
,
where (distance between slits), , and is the order of the maximum.

The angular width of the central maximum of single-slit diffraction is given by:
,
where is the slit width.

Given that 10 maxima of the double-slit pattern fit within the central maximum of the single-slit pattern, the angle for the 5th order (half of 10) double-slit maximum should equal the angle for the first minimum of the single-slit pattern:
.

Solving for :
.

This calculated width of the slit falls within the specified range. Therefore, the width of each slit is .

The central maximum shift indicates that the additional path difference introduced by the glass plate must correspond to a phase shift equivalent to four bright fringe separations. Let's calculate this thickness.
The phase difference caused by the glass plate is given by the formula:
where is the thickness of the glass plate, is the refractive index, and is the wavelength of light.
The central maximum shifts to the position of the 4\textsuperscript{th} bright fringe, introducing an additional path difference of . Thus, we equate and solve for :
Solving for :
Given and :
Now calculate:
This evaluates to approximately:
Thus, the thickness of the glass plate is 4 µm, which falls within the expected range of 4 to 4.

When light emerges out of a convex lens with a source placed at its focus, the rays emerge parallel to the principal axis. This results in a plane wavefront due to the parallel nature of the emergent rays.

Condition for Overlapping Fringes:
In Young’s double slit experiment, the condition for overlapping fringes for different wavelengths is given by:

where and are the fringe orders for wavelengths and , respectively.

Determine the Ratio of Wavelengths:
Given:

The ratio of the wavelengths is:

Find the Minimum Order of Overlapping Fringes:
For the fringes to overlap, .
Let and (the smallest integers that satisfy the ratio):

Conclusion:
The minimum order of fringe produced by which overlaps with the fringe produced by is .

In a double slit experiment, the condition for a dark fringe can be stated as:
d sin θ = (m + 0.5)λ.
Given that the dark fringe is observed at point P, we can approximate sin θ ≈ θ ≈ x/D, where x is the distance of the fringe from the center, and D is the distance to the screen.

Using the given data:
Wavelength λ = 400 nm = 400 × 10^-9 m,
Distance D = 0.2 m.

The first dark fringe corresponds to m = 0, so:
d(0) + 0.5(400 × 10^-9) = d.

Since x is approximately D:
d = λ/2.

Calculating:
d = (400 × 10^-9)/2 = 200 × 10^-9 m = 0.2 mm.

The minimum distance between the slits is 0.2 mm, which falls within the range [0.2, 0.2] mm.

In Young's double slit experiment, the intensity of light passing through a slit is directly proportional to the width of the slit.
Let the widths of the two slits be and . Given that the width of one slit is half the width of the other slit, let and .
The intensities of light from the two slits are proportional to their widths. Let the intensities be and . The maximum intensity in the interference pattern occurs when the waves from the two slits interfere constructively, and is given by: Substituting the values of and : The minimum intensity in the interference pattern occurs when the waves from the two slits interfere destructively, and is given by: Substituting the values of and : The ratio of the maximum to the minimum intensity is: So, the ratio is .

Let the intensities of the two monochromatic light beams be and .
Given that the ratio of their intensities is 1:9, we can write .
Let and . The maximum intensity in the interference pattern is given by: Substituting the values of and : The minimum intensity in the interference pattern is given by: Substituting the values of and : The ratio of the maximum to the minimum intensity is: So, the ratio of the intensities of maximum to minimum is 4:1.

The problem asks for the value of x, where xI represents the difference between the maximum and minimum intensities in an interference pattern formed by the superposition of two coherent monochromatic light beams with intensities 4I and 9I.

Concept Used:

When two coherent light beams with intensities and interfere, the resulting intensity at any point depends on the phase difference between the beams. The maximum and minimum possible intensities in the interference pattern are given by:

1. Maximum Intensity ( ): This occurs during constructive interference, when the waves are in phase. The formula is:

2. Minimum Intensity ( ): This occurs during destructive interference, when the waves are out of phase by radians. The formula is:

Step-by-Step Solution:

Step 1: Identify the intensities of the two coherent light beams.

The given intensities are:

Step 2: Calculate the maximum intensity ( ) in the interference pattern.

First, find the square roots of the intensities:

Now, substitute these into the formula for maximum intensity:

Step 3: Calculate the minimum intensity ( ) in the interference pattern.

Using the square roots of the intensities found in the previous step, substitute them into the formula for minimum intensity:

Step 4: Find the difference between the maximum and minimum intensities.

The difference is .

Final Computation & Result:

The problem states that the difference between the maximum and minimum intensities is equal to xI.

Comparing our calculated difference with the given expression:

By comparison, we find the value of x.

The value of x is 24.

Step 1: Understanding the fringe width formula. The fringe width in Young’s double-slit experiment is given by: where: - is the wavelength of the light, - is the distance between slits and screen, - is the separation between the slits.

Step 2: Analyzing Assertion (A). Since , red light ( is larger) produces wider fringes than blue light ( is smaller). Thus, Assertion (A) is incorrect because it states the opposite.