Diffraction and the Wave Theory of Light MCQ Quiz - Objective Question with Answer for Diffraction and the Wave Theory of Light - Download Free PDF

Calculation:

To determine the maximum number of intensity minima in the Fraunhofer diffraction pattern of a single slit, we use the condition for minima given by:

\(a \sin \theta = m \lambda\)

where;

a: is the slit width \(10 \, \mu \mathrm{m} = 10 \times 10^{-6} \, \mathrm{m}\)

\(\lambda\): is the wavelength of light \(0.630 \, \mu \mathrm{m} = 0.630 \times 10^{-6}m\).

m:  is the order of the minima \(m = \pm 1, \pm 2, \pm 3, \)

The maximum order m  can be found using:

\(m_{\text{max}} = \frac{a}{\lambda}.\)

Substituting the given values:

\(m_{\text{max}} = \frac{10 \times 10^{-6}}{0.630 \times 10^{-6}} \approx 15.87.\)

Since m must be an integer, the maximum order of minima is:

\(m = 15\)

Thus, option '4' is correct.

The correct answer is: Option 2) λ /√3 

Concepts:

In a single-slit diffraction pattern, the condition for maxima (bright fringes) can be derived from the interference of light waves. The angle of diffraction θ for the first maximum is given by:

sin(θ) = (m + 0.5) ×  λ / e, where m is the order of the maximum (m = 0 for first maximum), λ is the wavelength of light, and e is the slit width.

Calculation:

Given:

Wavelength of light, λ = 2

Angle of diffraction for the first maximum, θ = π/3

For the first maximum (m = 0):

sin(π/3) = (0 + 0.5) × λ  / e

√3/2 = 1/2 × λ / e

e = λ / √3.

Calculation:

In case of YDSE the distance of n^th maxima from central maxima is given by

\(\mathrm{Y}=\frac{\mathrm{n} λ \mathrm{D}}{\mathrm{~d}}\)

Here n , D & d are same

So, y × λ 

⇒ \(\frac{\mathrm{y}_{2}}{\mathrm{y}_{1}}=\frac{\lambda_{2}}{\lambda_{1}} \Rightarrow \frac{\mathrm{y}_{2}}{10 \mathrm{~mm}}=\frac{660 \mathrm{~nm}}{600 \mathrm{~nm}}\)

⇒ y₂ = 11 mm

Concept:

The diffraction pattern from a single slit can be analyzed using the formula for the angular width of the central maximum, which is given by \( \theta = \frac{\lambda}{a} \), where λ" id="MathJax-Element-428-Frame" role="presentation" style="position: relative;" tabindex="0">λλ" id="MathJax-Element-137-Frame" role="presentation" style="position: relative;" tabindex="0">λλ λ $\lambda$ is the wavelength of light and a" id="MathJax-Element-429-Frame" role="presentation" style="position: relative;" tabindex="0">aa" id="MathJax-Element-138-Frame" role="presentation" style="position: relative;" tabindex="0">aa a $a$ is the slit width.

Explanation: 

The linear width of the central maximum on the screen is given by \( 2L\tan(\theta) \approx 2L\theta \) when θ" id="MathJax-Element-430-Frame" role="presentation" style="position: relative;" tabindex="0">θθ" id="MathJax-Element-139-Frame" role="presentation" style="position: relative;" tabindex="0">θθ θ $\theta$ is small, where L" id="MathJax-Element-431-Frame" role="presentation" style="position: relative;" tabindex="0">LL" id="MathJax-Element-140-Frame" role="presentation" style="position: relative;" tabindex="0">LL L $L$ is the distance from the slit to the screen.

Given:

• Slit width, a=0.1mm=0.1×10−3m" id="MathJax-Element-432-Frame" role="presentation" style="position: relative;" tabindex="0">a=0.1mm=0.1×10−3ma=0.1mm=0.1×10−3m" id="MathJax-Element-141-Frame" role="presentation" style="position: relative;" tabindex="0">a=0.1mm=0.1×10−3ma=0.1mm=0.1×10−3m a=0.1mm=0.1×10−3m $a = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m}$
• Linear width of central maxima, w=5mm=5×10−3m" id="MathJax-Element-433-Frame" role="presentation" style="position: relative;" tabindex="0">w=5mm=5×10−3mw=5mm=5×10−3m" id="MathJax-Element-142-Frame" role="presentation" style="position: relative;" tabindex="0">w=5mm=5×10−3mw=5mm=5×10−3m w=5mm=5×10−3m $w = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m}$
• Distance from the slit to the screen, L=50cm=0.5m" id="MathJax-Element-434-Frame" role="presentation" style="position: relative;" tabindex="0">L=50cm=0.5mL=50cm=0.5m" id="MathJax-Element-143-Frame" role="presentation" style="position: relative;" tabindex="0">L=50cm=0.5mL=50cm=0.5m L=50cm=0.5m $L = 50 \, \text{cm} = 0.5 \, \text{m}$

Using the approximation w≈2Lθ" id="MathJax-Element-435-Frame" role="presentation" style="position: relative;" tabindex="0">w≈2Lθw≈2Lθ" id="MathJax-Element-144-Frame" role="presentation" style="position: relative;" tabindex="0">w≈2Lθw≈2Lθ w≈2Lθ $w \approx 2L\theta$ :

\( \theta = \frac{w}{2L} = \frac{5 \times 10^{-3}}{2 \times 0.5} = 5 \times 10^{-3} \, \text{radians} \)

Now, using the formula θ=λa" id="MathJax-Element-436-Frame" role="presentation" style="position: relative;" tabindex="0">θ=λaθ=λa" id="MathJax-Element-145-Frame" role="presentation" style="position: relative;" tabindex="0">θ=λaθ=λa θ=λa $\theta = \frac{\lambda}{a}$ :

\( \lambda = \theta \times a = 5 \times 10^{-3} \times 0.1 \times 10^{-3} = 5 \times 10^{-7} \, \text{m} \)

The correct option is (3).

Calculation:
1 / λ = RZ² [(1 / n_f²) - (1 / n_i²)]

Thus,

λ₂ / λ₁ = [(1 / 9) - (1 / n²)] / (8 / 9)

Here n is the principal quantum number of the initial excited state.

Angular width: θ = 2λ / d

θ₂ / θ₁ = λ₂ / λ₁ = 2 / 25 = (n² - 9) / (8n²)

Thus, we get n = 5

Now, θ₁ = 2λ₁ / d

1 / λ₁ = 2 / (d × θ₁) = 2 / (0.01 × 10^-3 × 6.4 × 10^-2) = 2 / (6.4 × 10⁷) m

But, 1 / λ₁ = RZ² × [(1 / 9) - (1 / n²)]

Thus, 2 / (6.4 × 10⁷) = 1.097 × 10⁷ Z² × [(1 / 9) - (1 / 25)]

On solving, we get Z = 2

Explanation:

• A corona is an optical phenomenon produced by the diffraction of sunlight or moonlight by individual small water droplets and sometimes tiny ice crystals of a cloud.
• Corona consists of several concentric, pastel-colored rings around the celestial object and a central bright area called aureole.
• The aureole is often (especially in the case of the Moon) the only visible part of the corona and has the appearance of a bluish-white disk which fades to reddish-brown towards the edge.
• The angular diameter of a corona depends on the sizes of the water droplets involved; smaller droplets produce larger corona.

The correct answer is black.

• The colour of the sky is black that an astronaut will see in space.
• Because of the scattering of light particles, the sky appears blue when viewed from the earth.

Key Points

• But when there is an atmosphere, blue light gets refracted.
  • The actual colour of the sky is black.
  • And there is no atmosphere in the space.
  • So the astronauts see the black colour of the sky.
  • There is no scattered light to reach our eyes in outer space.
  • A range of wavelengths of light is emitted by the sky.
  • All the wavelength of light can’t be seen by us.

In diffraction

d sin 30° = λ

\( \Rightarrow \lambda = \frac{d}{2}\)

Now, let x is the separation between two slits.

Young’s fringe width (ω) = 1 cm = 10^-2 m

\(\omega = \frac{{\lambda D}}{x}\)

\({10^{ - 2}} = \frac{{d \times 50 \times {{10}^{ - 2}}}}{{2x}}\)

Bragg’s Law:

Bragg's Law relates the angle θ (at which there is a maximum in diffracted intensity) to the wavelength of X-rays and the interlayer distance d between the planes of atoms/ions/molecules in the lattice.

Condition for constructive interference (maxima in the reflected).

nλ = 2d sinθ

λ = Wavelength of the X-ray

d = distance of crystal layers

θ = incident angle (the angle between the incident plane and the scattering plane)

n = integer (order)

This condition is known as Bragg’s law.

From the above equation, 

\(λ = \frac {2d sin θ}{n}\)

Calculation:

Given d = 0.282 nm, θ = 7°, n = 2; 

From the above relation,

\(\lambda = \frac{{2dsinθ }}{2} = dsinθ\)

The correct answer is Less than the angle of incidence.

Key Points

• The cause of refraction is a change in light speed and wherever the light speed changes most, the refraction is greatest.
• The speed is related to the optical density of a material that is related to the index of refraction of a material.
• Air is the least dense material (lowest index of refraction value) and water is the denser material (largest index of refraction value).
• Thus, it would be reasonable that the most refraction occurs for the transmission of light across an air-water boundary.
• Hence, the angle of refraction will be smaller than the angle of incidence.

Important Points

• Refraction:
  • ​When a ray of light passes from one medium to another it suffers a change in direction at the boundary of two mediam is called refraction.
  • The change in direction because the speed of light travels at different speeds in different mediums.
  • The Refractive Index of a medium defines that it is denser or rarer.

​ 

Concept:

Diffraction is the phenomenon when passes through space or hole it tends to spread out, this phenomenon of waves is known as diffraction.

• The diffraction from the slit of the first maximum to the central maximum is the function of sinusoidal waves which can be expressed as 

The intensity is given by :

\(I=I_0 (\frac{sin(\phi/2)}{\phi/2})\)

For nth second maxima, dsinθ = (2n+1)/2\(\lambda\)

So, \(\frac{\phi}{2} = \frac{\pi}{2}[d sin\theta] = \frac{2n+1}{2}\lambda\)

So, \(I=I_o(\frac{sin(2n+1)\lambda/2}{(2n+1)\lambda/2})^2\)

\(I=\frac{I_o}{{(2n+1)\lambda/2})^2}\)

So \(I_0:I_1 = I_o: \frac{4I_o}{9\pi^2}\)

\(I_0:I_1 =1:0.045\)

\(I_0:I_1 =1000:45 = 200:9\)

Note: This question was deleted in RSMSSB Lab Assistant 2016 Exam

Concept:

The path difference is the difference in the distance travelled by two beams when they are scattered in the same direction from different points. It is equal to d sinθ

Constructive interference (Bright Fringe):

d sinθ = nλ

n = 0: Central (bright) fringe

n = 1: First bright fringe

Destructive interference:

\(d\sin \theta = \left( {n + \frac{1}{2}} \right)λ\)

n = 0: First dark fringe (next to the central fringe)

n = 1: second dark fringe

The separation on the screen between the centres of two consecutive bright fringes or two consecutive dark fringes is called fringe width.

The centres of the bright fringes are obtained at y from the central fringe:

\(y = \frac{D}{d}nλ\)

The centres of the dark fringes are obtained at y from the central fringe:

\(y = \frac{D}{d}\left( {n + \frac{1}{2}} \right)λ\)

The fringe width:

\(w = \frac{D}{d}λ\)

D is the separation between the slits and the screen and d is the separation between the slits.

• The central fringe is n = 0.
• The fringe to either side of the central fringe has an order of n = 1 (the first order fringe).
• The order of the next fringe out on either side is n = 2 (the second-order fringe).

Calculation:

λ = 5800 Å = 5800 × 10^-10 m = 5800 × 10^-7 mm

d = 0.2 mm, D = 2 m = 2000 mm

\( y = \frac{D}{d}\left( {n + \frac{1}{2}} \right)λ \)

\(y = \frac{D}{d} \times \frac{1}{2} \times λ \)

\(= \frac{{2000}}{{0.2}} \times \frac{1}{2} \times 5800 \times {10^{ - 7}}\)

\( λ= 2.9mm \)

The distance of the first dark fringe on either side will be 2 times λ , i.e.

Concept:

• The x-rays are high energy electromagnetic radiation. They have energies ranging from about 200 eV to 1 MeV which puts them between γ-rays and UV rays.
• γ-rays and x-rays are essentially identical, γ-rays being somewhat more energetic and shorter in wavelength than x-rays.
• The wavelength of X-ray photons is on the order of the distance between atomic nuclei in solids, e.g. 4 angstroms (bonds are roughly 1.5-2.5 Å). This shows that the waves are fit nice and snugly between the atoms and "fill" the crystal and thereby give us information about where the "cavities" are and with the help of cavities, we can determine where the "rigid" stuff is (the atoms).
• Gamma rays are about 100 times smaller in wavelength than X-rays (so instead of 4 Å, something like 0.04 Å, and so they generally penetrate through any solid material and don't produce a clear diffraction pattern like X-rays. Also it is much more difficult to produce γ-rays.

The Angstrom (Å) is a unit of length equal to 10^-10 m. The Angstrom was widely used as a unit of wavelength for electromagnetic radiation covering the visible part of the electromagnetic spectrum and x-rays.This unit is also used for interatomic spacings. 

Concept:

A diffraction grating consists of a large number (n) of equally spaced narrow slits or lines.

d sin θ = mλ

Calculation:

Given:

Let there are n lines over 3 cm wide diffraction grating.

d = 3/n cm = 30/n mm

λ = 600 nm = 600 × 10^-9 m, d = 3 cm = 30 mm,

d sin θ = mλ

\(\frac{{30}}{n} \times \sin 30^\circ = 2 \times 600 \times {10^{ - 6}}\)

CONCEPT:

Diffraction​​:

• ​​​It is defined as the bending of waves around the corners of an obstacle​.
• The condition of diffraction is that the width of the obstacle must be less than or comparable with the wavelength of the wave.
• Diffraction is the characteristic of all types of waves.
• The greater the wavelength of the wave higher will be its degree of diffraction.
• The condition of diffraction is that the width of the obstacle must be less than or comparable with the wavelength of the wave.

EXPLANATION:

• Diffraction is the characteristic of all types of waves. i.e., the greater the wavelength of the wave higher will be its degree of diffraction.
• Since the wavelength of the microwaves is maximum, so the microwaves will show maximum diffraction.
• Hence, option 3 is correct.