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 physics optoelectronic devices resistance and temperature

Students in a lab are trying to determine the wavelength of laser light by shining the laser through two narrow slits separated by $9.60$ micrometers toward a screen $4.00$ meters away. The first order bright fringes are $23.0$ centimeters from the center fringe.
What wavelength do the students determine?

  1. $680 nm$
  2. $552 nm$
  3. $883 nm$
  4. $167 nm$
  5. $8,832 nm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given : $d = 9.6$ $\mu m = 9.6 \times 10^{-6}$ $m$      $D = 4.00$  $m$       $y _1 = 23$ cm $ = 0.23$ m

Distance of first order bright fringe from central fringe, $y _1 = \dfrac{\lambda D}{d}$
$\therefore$   $0.23 = \dfrac{\lambda \times 4.00}{9.6 \times 10^{-6}}$                      

$\implies \lambda = 0.552 \times 10^{-6}$ m $ =552$  $nm$

Multiple choice physics wave optics interference

Two points P and Q are situated at the same distance from a source of light but on opposite sides.The phase difference between the light waves passing through P and Q will be

  1. $\pi$
  2. 2$\pi$
  3. $\pi$/2

  4. 0
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If two points are at the same distance from a source, the light waves arrive at those points with the same path length. Therefore, the path difference is 0, which results in a phase difference of 0.

Multiple choice physics wave optics interference

In an interference experiment, distance between the lists is $2\ mm$ and screen is placed at distance $1\ m$ from the slits. Fourth dark fringe is formed exactly opposite to one of the slits. Wavelength of light used in nm is

  1. 480

  2. 600

  3. 570

  4. 500
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics wave optics interference

In YDSE $S _1$ and $S _2$ has intersity $I$ and $9I$. Find difference in intensity b/w point which has phase difference of  $\pi$

  1. $10 I$
  2. $6 I$
  3. $8I$
  4. $4I$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Resultant intensity I_res = I1 + I2 + 2 * sqrt(I1 * I2) * cos(phi). Given I1 = I, I2 = 9I, phi = pi. cos(pi) = -1. I_res = I + 9I + 2 * sqrt(9 * I^2) * (-1) = 10I - 6I = 4I.

Multiple choice physics wave optics interference

The path difference between two interfering waves at a point on the screen  is $ \lambda /6 $. The ratio of intensity at the point and that the central bright fringe will be (Assume that internally due to each slit in same). 

  1. 0.853

  2. 8.53

  3. 0.75

  4. 7.5
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that,

Path difference $x=\dfrac{\lambda }{6}$

We know that,

  $ I={{I} _{0}}{{\cos }^{2}}\left( \dfrac{\phi }{2} \right) $

 $ \dfrac{I}{{{I} _{0}}}={{\cos }^{2}}\left( \dfrac{\phi }{2} \right)....(I) $

Now, the phase difference is

  $ \phi =\dfrac{2\pi }{\lambda }\times x $

 $ \phi =\dfrac{2\pi }{\lambda }\times \dfrac{\lambda }{6} $

 $ \phi =\dfrac{\pi }{3} $

Now, put the value of $\phi $ in equation (I)

  $ \dfrac{I}{{{I} _{0}}}={{\cos }^{2}}{{30}^{0}} $

 $ \dfrac{I}{{{I} _{0}}}=\dfrac{3}{4} $

 $ \dfrac{I}{{{I} _{0}}}=0.75 $

Hence, the value of ratio of the intensity at the point is $0.75$ 

Multiple choice physics wave optics interference

The distance between the two slits in a Young's double slit experiment is $d$ and the distance of the screen from the plane of the slits is $b$,$P$ is a point on the screen directly in front of one of the slits. The path difference between the waves arriving at $P$ from the two slits is

  1. $\dfrac{d^{2}}{b}$
  2. $\dfrac{d^{2}}{2b}$
  3. $\dfrac{2d^{2}}{b}$
  4. $\dfrac{d^{2}}{4b}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The path difference at a point directly in front of one slit is the difference in distances from the two slits. Using the Pythagorean theorem, the distance from the far slit is sqrt(b^2 + d^2) and from the near slit is b. Path difference = sqrt(b^2 + d^2) - b. Using binomial expansion for d << b, sqrt(b^2 + d^2) - b = b(1 + d^2/b^2)^(1/2) - b approx b(1 + d^2/2b^2) - b = d^2 / 2b.

Multiple choice physics wave optics interference

In a YDSE, the central bright fringe can be identified :

  1. as it has greater intensity than the other bright fringe.

  2. as it is wider than the other bright fringes.

  3. as it is narrower than the other bright fringes.

  4. by using white light instead of single wavelength light.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In a YDSE, the central bright fringe occurs where the path difference is zero for all wavelengths, resulting in maximum constructive interference and thus the highest intensity.

Multiple choice physics wave optics interference

Two coherent plane light waves of equal amplitude makes a small angle $\alpha (<<1)$ with each other. They fall almost normally on a screen. If $\gamma $ is the wavelength of light waves, the fringe width $\Delta x$ of interference patterns of the two sets of wave on the screen is  

  1. $\dfrac { 2\lambda }{ \alpha } $
  2. $\dfrac { \lambda }{ \alpha } $
  3. $\dfrac { \lambda }{ (2\alpha ) } $
  4. $\dfrac { \lambda }{ \sqrt { \alpha } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When two plane waves make a small angle alpha, the fringe width is given by beta = lambda / alpha. This is derived from the geometry of the interference pattern formed by the two waves.

Multiple choice physics wave optics interference

The path difference between two wavefronts emitted by coherent sources of wavelength 5460 $\overset{o}{A}$ is 2.1 micron. The phase difference between the wavefronts at that point is

  1. 7.962
  2. 7.962 $\pi$
  3. $\displaystyle\frac{7.962}{\pi}$
  4. $\displaystyle\frac{7.962}{3\pi}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Phase diff. = $\displaystyle\frac{2\pi x}{\lambda}$
Path difference = $\displaystyle\frac{2\pi \times 2.1 \times 10^{-6}}{5460 \times 10^{-10}}$ = 7.692 $\pi$ radian.

Multiple choice physics wave optics interference

Two light rays having the same wavelength $\lambda$ in vacuum are in phase initially. Then the first ray travels a path ${L} _{1}$ through a medium of refractive index ${n} _{1}$ while the second ray travels a path of length ${L} _{2}$ through a medium of refractive index ${n} _{2}$. The two waves are then combined to produce interference. The phase difference between the two waves is:

  1. $\dfrac { 2\pi }{ \lambda } \left( { L } _{ 2 }-{ L } _{ 1 } \right) $
  2. $\dfrac { 2\pi }{ \lambda } \left( { n } _{ 1 }{ L } _{ 1 }-{ n } _{ 2 }{ L } _{ 2 } \right) $
  3. $\dfrac { 2\pi }{ \lambda } \left( { n } _{ 2 }{ L } _{ 1 }-{ n } _{ 1 }{ L } _{ 2 } \right) $
  4. $\dfrac { 2\pi }{ \lambda } \left( \dfrac { { L } _{ 1 } }{ { n } _{ 1 } } -\dfrac { { L } _{ 2 } }{ { n } _{ 2 } } \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The optical path between any two points is proportional to the time of travel.
The distance traversed by light in a medium of refractive index $\mu $ in time $t$ is given by
$d=vt$            .....(i)


where $v$ is velocity of light in the medium. The distance traversed by light in a vacuum in this time,

$\Delta =ct$

  $=c\cdot \dfrac { d }{ v } $        [from equation (i)]

  $=d \dfrac { c }{ v } =\mu d$          .......(ii)                   (Since, $\mu =\dfrac { c }{ v } $)

This distance is the equivalent distance in vacuum and is called optical path.

Here, optical path for first ray $={ n } _{ 1 }{ L } _{ 1 }$

Optical path for second ray $={ n } _{ 2 }{ L } _{ 2 }$

Path difference $={ n } _{ 1 }{ L } _{ 1 }-{ n } _{ 2 }{ L } _{ 2 }$

Now, phase difference

    $=\dfrac { 2\pi  }{ \lambda  } \times $ path difference

    $=\dfrac { 2\pi  }{ \lambda  } \times \left( { n } _{ 1 }{ L } _{ 1 }-{ n } _{ 2 }{ L } _{ 1 } \right) $

Multiple choice physics oscillations and waves huygen's wave theory refraction of water waves reflection and refraction at plane surfaces theories on light

In frounhofer diffraction by a single slit, a position where first order minimum is formed by the wave length $6000\mathring { A } $, first order maximum is formed due to unknown wavelength, the unknown wavelength is

  1. $5000\mathring { A } $
  2. $6000\mathring { A } $
  3. $4000\mathring { A } $
  4. $9000\mathring { A } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In single-slit diffraction, the condition for minima is a*sin(theta) = n*lambda. For the first order minimum, a*sin(theta) = 1 * 6000 A. For the first order maximum, the condition is a*sin(theta) = (n + 1/2) * lambda_unknown. Setting these equal, 6000 = 1.5 * lambda_unknown, so lambda_unknown = 4000 A.

Multiple choice physics superposition and interference of sound waves

Two coherent sources of intensity ratio $\alpha$ interfere. In interference pattern $\dfrac{{I} _{max} - {I} _{min}}{{I} _{max} + {I} _{min}} =$

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

$\dfrac { { I } _{ max }-{ I } _{ min } }{ { I } _{ max }+{ I } _{ min } } =\dfrac { { \left( { a } _{ 1 }+{ a } _{ 2 } \right)  }^{ 2 }-{ \left( { a } _{ 1 }-{ a } _{ 2 } \right)  }^{ 2 } }{ { \left( { a } _{ 1 }+{ a } _{ 2 } \right)  }^{ 2 }+{ \left( { a } _{ 1 }-{ a } _{ 2 } \right)  }^{ 2 } }$
                                       [$\because { I } _{ max }={ \left( { a } _{ 1 }+{ a } _{ 2 } \right)  }^{ 2 },{ I } _{ min }={ \left( { a } _{ 1 }-{ a } _{ 2 } \right)  }^{ 2 }$  where $a =$ amplitude]
                $=\dfrac { 4{ a } _{ 1 }{ a } _{ 2 } }{ 2\left( { a } _{ 1 }^{ 2 }+{ a } _{ 2 }^{ 2 } \right)  } =\dfrac { 2{ a } _{ 1 }{ a } _{ 2 } }{ { a } _{ 1 }^{ 2 }+{ a } _{ 2 }^{ 2 } } $
Now, dividing the numerator and denominator by ${a} _{1}{a} _{2}$, we get
$\dfrac { { I } _{ max }-{ I } _{ min } }{ { I } _{ max }+{ I } _{ min } } =\dfrac { 2 }{ \left[ \dfrac { { a } _{ 1 } }{ { a } _{ 2 } } +\dfrac { { a } _{ 2 } }{ { a } _{ 1 } }  \right]  } =\dfrac { 2 }{ \left[ \sqrt { \alpha  } +\dfrac { 1 }{ \sqrt { \alpha  }  }  \right]  } =\dfrac { 2\sqrt { \alpha  }  }{ \left( \alpha +1 \right)  } $

Multiple choice standing waves waves physics

Consider single slit experiment of diffraction of light. If light of wavelength $5000\ \mathring { A } $ fall on a slit of width $1\mu\ m$ then the angular width of central maximum. 

  1. ${30}^{o}$
  2. ${15}^{o}$
  3. ${60}^{o}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics superposition of waves-1: interference and beats distinction between interference and beats beats and its applications beats in sound waves

Two monochromatic light waves of amplitudes $A$ and $2A$ interfering at a point, have a phase difference of ${60^0}.$ The intensity at that point will be  proportional to :

  1. $3{A^2}$
  2. $5{A^2}$
  3. $7{A^2}$
  4. $9{A^2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The resultant intensity I is given by I = I1 + I2 + 2 * sqrt(I1 * I2) * cos(phi). Since intensity is proportional to amplitude squared, I1 = k * A^2 and I2 = k * (2A)^2 = 4 * k * A^2. With phi = 60 degrees, cos(60) = 0.5. Thus, I = k * A^2 + 4 * k * A^2 + 2 * sqrt(k * A^2 * 4 * k * A^2) * 0.5 = 5 * k * A^2 + 2 * (2 * k * A^2) * 0.5 = 7 * k * A^2.

Multiple choice physics light and shadow formation of image by a pinhole camera pinhole camera shadow

Intensity of light due to a point source at a distance $2m$ is $20$ units. The intensity due to the same source at a distance of $4m$ is

  1. $2.5\ units$
  2. $6\ units$
  3. $80\ units$
  4. $5\ units$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Intensity follows the inverse square law, I proportional to 1/r^2. If distance doubles from 2m to 4m, intensity becomes 1/4 of the original value: 20 / 4 = 5 units.