Physics

Wave Optics and Diffraction

148 Questions

Wave optics explores light phenomena like interference and diffraction. This hub features problems on Young's double slit experiment, Fresnel biprism, and single slit diffraction patterns. These topics regularly appear in physics sections of competitive exams.

Young's double slitFresnel biprismSingle slit diffractionConstructive interferenceFringe width calculations

Wave Optics and Diffraction Questions

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In a Young's double slit experiment the intensity at a point where the path difference is $ \dfrac { \lambda  }{ 6 } $ ($\lambda $ being the wavelength of the light used) is $I$. if ${ I } _{ 0 }$ denotes the maximum intensity, is equal to

  1. $\dfrac { 1 }{ \sqrt { 2 } } $
  2. $\dfrac { \sqrt { 3 } }{ 2 } $
  3. $\dfrac { 1 }{ 2 } $
  4. $\dfrac { 3 }{ 4 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For maximum intensity $\phi =0$

Assume that ${ I } _{ 1 }={ I } _{ 2 }={ I } _{ 3 }$(source intensity)

$\therefore { I } _{ 0 }={ I } _{ 3 }+{ I } _{ 2 }+2\sqrt { { I } _{ 2 }{ I } _{ 3 } } \\ { I } _{ 0 }={ 4I } _{ 3 }\longrightarrow 1\\ \Delta x=\cfrac { \lambda  }{ 6 } \\ \therefore \Delta \phi =\cfrac { 2\pi  }{ \lambda  } \Delta x\quad =\cfrac { 2x }{ \lambda  } \times \cfrac { \lambda  }{ 6 } \\ \quad \quad \quad \quad =\cfrac { \pi  }{ 3 } \\ \therefore I={ I } _{ s }+{ I } _{ s }+2\sqrt { { I } _{ s }{ I } _{ s } } \cos { \cfrac { \pi  }{ 3 }  } \\ \quad \quad =3{ I } _{ 3 }\\ I=\cfrac { 3 }{ 4 } { I } _{ 0 }$

 

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In Young's double slit experiment using monochromatic light of wavelengths $\lambda$, the intensity of light at a point on the screen with path difference $\lambda$ is M units. The intensity of light at a point where path difference is $\lambda/3$ is then

  1. $\frac{M}{2}$
  2. $\frac{M}{4}$
  3. $\frac{M}{8}$
  4. $\frac{M}{16}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Resultant intensity
$I _R = I _1 + I _2 + 2\sqrt{I _1 I _2} cos \theta$
If path difference = $\lambda$ phase difference $2 \pi$
$\therefore I _R = I + I + 2 \sqrt{I \times I} cos 2 \pi = 4 I = M$
If path difference $\dfrac{\lambda}{3}$, phase difference $\phi = \dfrac{2 \pi}{3} rad$
$I _R' = I + I + 2 \sqrt{I \times I} cos \dfrac{2 \pi}{3} = I = \dfrac{M}{4}$

Multiple choice

In Young's double slit experiment, the brightest lines observed were caused by:

  1. Interference of waves in phase

  2. Interference of waves out of phase

  3. The light reaching this area first

  4. This area containing specific elements
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In Young's double slit experiment, bright fringes occur where light waves from the two slits arrive in phase, leading to constructive interference. Dark fringes occur where waves arrive out of phase, leading to destructive interference.

Multiple choice physics mechanical properties of solids applications of elasticity elastic energy properties of matter

In the Young's double slit experiment the intestines at two points $P _{1}$ and $P _{2}$ on the screen are respectively $I _{1}$ and $I _{2}$. If $P _{1}$ is located at the centre of bright fringe and $P _{2}$ is located at a distance equal to a quarter of fringe width from $P _{1}$, then $I _{1}/I _{2}$ is 

  1. $2$
  2. $1/2$
  3. $4$
  4. $16$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Intensity I = 4 * I0 * cos^2(phi/2). At the center (P1), phi = 0, so I1 = 4 * I0. At a distance of quarter fringe width (P2), the path difference is lambda/4, so phase difference phi = 2 * pi * (lambda/4) / lambda = pi/2. I2 = 4 * I0 * cos^2(pi/4) = 4 * I0 * (1/2) = 2 * I0. Therefore, I1/I2 = 4 * I0 / 2 * I0 = 2.

Multiple choice physics electromagnetic waves properties and uses of different radiation electromagnetic spectrum spectrum

Maximum diffraction takes place in a given slit for

  1. $\gamma $-rays

  2. Ultraviolet light

  3. Infrared light

  4. Radio waves
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
When the gap size is larger than the wavelength, the wave passes through the gap and does not spread out much on the other side. When the gap size is equal to the wavelength, maximum diffraction occurs. So among all of them radio waves have minimum wavelength and hence it will give maximum diffraction. 
Multiple choice

In the double-slit experiment, what happens when a single electron passes through both slits at the same time?

  1. It creates an interference pattern.

  2. It behaves like a classical particle.

  3. It disappears and reappears on the other side.

  4. It splits into two electrons.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In the double-slit experiment, when a single electron passes through both slits at the same time, it creates an interference pattern on the screen behind the slits. This pattern is evidence of the wave-like nature of electrons and demonstrates the principle of quantum superposition.

Multiple choice

In the double-slit experiment, when a beam of electrons is passed through two closely spaced slits, the electrons:

  1. Behave like particles and create two distinct bands on a screen

  2. Behave like waves and create an interference pattern on a screen

  3. Behave randomly and create a scattered pattern on a screen

  4. Behave chaotically and create an unpredictable pattern on a screen
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In the double-slit experiment, when a beam of electrons is passed through two closely spaced slits, the electrons behave like waves and create an interference pattern on a screen.

Multiple choice

In the double-slit experiment, what happens when a single electron passes through both slits?

  1. It creates two bright spots on the screen.

  2. It creates one bright spot on the screen.

  3. It creates an interference pattern on the screen.

  4. It disappears.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When a single electron passes through both slits, it creates an interference pattern on the screen, indicating that it behaved like a wave.

Multiple choice

What is the role of fringe spacing in interferometry and how does it relate to the baseline length?

  1. It determines the angular resolution of the observations

  2. It affects the sensitivity of the radio telescopes

  3. It is related to the wavelength of the radio waves

  4. It influences the data processing requirements
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Fringe spacing in interferometry is the distance between adjacent fringes in the interference pattern. It is inversely proportional to the baseline length, and a smaller fringe spacing corresponds to higher angular resolution, enabling astronomers to resolve finer details in the observed astronomical objects.

Multiple choice

What is the condition for constructive interference?

  1. The path difference between the waves is an integer multiple of the wavelength.

  2. The path difference between the waves is a half-integer multiple of the wavelength.

  3. The path difference between the waves is equal to the wavelength.

  4. The path difference between the waves is equal to half the wavelength.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Constructive interference occurs when the path difference between the waves is an integer multiple of the wavelength, resulting in the waves reinforcing each other.

Multiple choice

What is the relationship between the wavelength of a wave and the size of the interference or diffraction pattern?

  1. The wavelength is directly proportional to the size of the pattern.

  2. The wavelength is inversely proportional to the size of the pattern.

  3. The wavelength is not related to the size of the pattern.

  4. The relationship between the wavelength and the size of the pattern depends on the specific experiment.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The wavelength of a wave is inversely proportional to the size of the interference or diffraction pattern, meaning that shorter wavelengths produce smaller patterns and longer wavelengths produce larger patterns.

Multiple choice

What is the name of the principle that states that the diffraction pattern produced by a single slit is the same as the interference pattern produced by two slits separated by a distance equal to the width of the single slit?

  1. Fermat's principle

  2. Huygens' principle

  3. Young's principle

  4. Fresnel's principle
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Fresnel's principle states that the diffraction pattern produced by a single slit is the same as the interference pattern produced by two slits separated by a distance equal to the width of the single slit.

Multiple choice

What is the Fraunhofer pattern in the Josephson effect?

  1. A pattern of interference fringes that is observed when a Josephson junction is illuminated with microwaves.

  2. A pattern of magnetic flux lines that is observed when a Josephson junction is subjected to a magnetic field.

  3. A pattern of current-voltage characteristics that is observed when a Josephson junction is biased with a voltage.

  4. A pattern of critical currents that is observed when a Josephson junction is subjected to a magnetic field.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Fraunhofer pattern in the Josephson effect is a pattern of interference fringes that is observed when a Josephson junction is illuminated with microwaves. This pattern is caused by the interference of the Cooper pairs that tunnel through the insulating layer.