Tag: wave optics

Quizzes (12) Questions (312)

Questions Related to wave optics

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

Light waves of wave length $\lambda$ propagate in a medium. If $M$ and $N$ are two points on the wave front and they are separated by a distance $\lambda /4$, the phase difference between them will be (in radian)

  1. $\dfrac{\pi}{2}$
  2. $\dfrac{\pi}{8}$
  3. $\dfrac{\pi}{45}$

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

Points on the same wavefront are in the same phase. Therefore, the phase difference between any two points on the same wavefront is zero, regardless of the distance between them.

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

A glass wedge of angle $0.01$ radian and $\mu=1.5$ is illuminated by monochromatic light of wavelength $6000\ A$ falling normally on it. At what distance from wedge will $10^{th}$ dark fringe be observed by reflected light ?

  1. $0.1\ mm$
  2. $0.2\ mm$
  3. $0.3\ mm$
  4. $0.4\ mm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The distance x of the nth dark fringe in a wedge-shaped film is given by x = n * lambda / (2 * mu * theta). Substituting the given values for wavelength, refractive index, and wedge angle gives 0.2 mm.

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

For interference between waves from two sources of intensities $I$ and $4I$, find the intensity at the point in the pattern where the phase difference is $\dfrac{\pi}{2}$ and $\pi$.

  1. $10I$ and $I$
  2. $5I$ and $5I$
  3. $5I$ and $I$
  4. $5I$ and $10I$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

I_res = I1 + I2 + 2 * sqrt(I1 * I2) * cos(phi). I1 = I, I2 = 4I. For phi = pi/2, cos(pi/2) = 0, so I_res = I + 4I = 5I. For phi = pi, cos(pi) = -1, so I_res = I + 4I + 2 * sqrt(4 * I^2) * (-1) = 5I - 4I = I.

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

The phase difference between two waves, represented by
${ y } _{ 1 }={ 10 }^{ -6 }sin{ 100t+(x/50)+0.5} m$
${ y } _{ 2 }={ 10 }^{ -6 }cos{ 100t+\left( \frac { x }{ 50 }  \right) } m$
where x is expressed in meters and is expressed in seconds, is approximately:

  1. 2.07 Radians

  2. 0.5 Radians

  3. 1.5 Radians

  4. 1.07 Radians
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

y1 = 10^-6 sin(100t + x/50 + 0.5). y2 = 10^-6 cos(100t + x/50) = 10^-6 sin(100t + x/50 + pi/2). The phase of y1 is phi1 = 100t + x/50 + 0.5. The phase of y2 is phi2 = 100t + x/50 + pi/2. Phase difference = |phi2 - phi1| = |pi/2 - 0.5| = |1.57 - 0.5| = 1.07 radians.

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.