Tag: diffraction

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Questions Related to diffraction

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

When a certain photosensistive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photo current is-${ V } _{ 0 }/2.$ When the surface frequency v/2, the stopping potential is -${ V } _{ 0 }.$ The threshold frequency for photoelectric emission is

  1. $\dfrac { 5v }{ 3 } $
  2. $2v$
  3. $\dfrac { 4 }{ 3 } v$
  4. $\dfrac { 3v }{ 2 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

A thin water of $9.9\mu$ m thickness and refractive index $1.5$ is introduced in front of one of the slits in a Young's double slit experiment. By how many fringe width does the fringe pattern shift? (Given wavelength of the wave is $5.5\times 10^{-7}$m)

  1. $11$
  2. $9$
  3. $3$
  4. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Shift in fringes N = (mu - 1) * t / lambda. N = (1.5 - 1) * 9.9 * 10^-6 / 5.5 * 10^-7 = 0.5 * 9.9 * 10^-6 / 5.5 * 10^-7 = 0.5 * 18 = 9.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

Monochromatic light of wavelength $400\ nm$ and $560\ nm$ are incident simultaneously and normally on double slits apparatus whose slits separation is $0.1\ mm$ and screen distance is $1\ m$. Distance between areas of total darkness will be:

  1. $4\ mm$
  2. $5.6\ mm$
  3. $14\ mm$
  4. $28\ mm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For $\lambda =400$mm

fringe width
$\beta =\cfrac { \lambda D }{ d } \ =\cfrac { 400\times 1\times 10\times 1000 }{ 1\times { 10 }^{ 6 } } $
$ =4$mm
For $\lambda =560$mm
$\beta =\cfrac { \lambda D }{ d } =5.6$mm
$LCM$ of both fringe width will give combined fringe width which is also the distance between $2$ minima .
$LCM 4,5,6=28$mm

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In, which of the following the interference is produced by the division of wave front? 

  1. Young's double slit experiment

  2. Fresnel's bi-prism

  3. Lloyd's mirror

  4. Colours of thin film
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In Young's Double Slit Experiment, wavefront from a single sources divides when passed through the two slits. These two wavefronts interfere after emerging out of the slits.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction
Monochromatic light beam passed through two narrow slits form a pattern of bright and dark lines on a screen. The phenomena causing this pattern is: 

  1. Refraction

  2. Reflection

  3. Polarization

  4. Interference

  5. Intensity
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The waves pass through the double-slit and interfere with each other, creating nodes and antinodes on the screen, which we see as alternating bright and dark lines. 

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

State whether given statement is True or False
In Young's double slit experiment, the two coherent sources are real.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
False.
Young's double slit experiment was done to demonstrate interference. The essential and the most sufficient condition for interference is that there must be a constant phase difference between the two waves emitted by the two sources (real or virtual). Such sources, having a constant phase difference and same frequency, are called coherent sources. Therefore, the sources used should be coherent, real or virtual does not matter. 
Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In Young's experiment, the source is red light of wavelength $7\times 10^{-7}m$. When a thin glass plate of refractive index $1.5$ at this wavelength is put is the path of one of the interfering beam, the central bright fringe shifts by $10^{-3} m$ to the position previously occupied by the $5^{th}$ bright fringe. Find the thickness of the plate. When the source is now changed to green light of wavelength $5\times 10^{-7}m$ the central fringe shifts to a position initially occupied by the $6^{th}$ bright fringe due to red light. Find the refractive index of glass for the green light. Also estimate the change in fringe width due to the change in wavelength.

  1. $7\times 10^{-6} m, 1.6, 5.7\times 10^{-5}m$.
  2. $8\times 10^{-6} m, 1.6, 5.7\times 10^{-5}m$.
  3. $9\times 10^{-6} m, 1.6, 5.7\times 10^{-5}m$.
  4. $17\times 10^{-6} m, 1.6, 5.7\times 10^{-5}m$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The shift x = (mu - 1) * t * D / d. For red light, x = 5 * beta_red = 5 * lambda_red * D / d. Given x = 10^-3 m, lambda_red = 7 * 10^-7 m, we find t. The calculations for thickness, refractive index, and change in fringe width are complex but consistent with option A.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

If the intensity of the waves observed by two coherent sources is $I$. Then the intensity if resultant wave in constructive interference will be:-

  1. $2 I$
  2. $4 I$
  3. $I$

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

During constructive interference, the amplitude of the resultant wave is the sum of the individual amplitudes (a + a = 2a). Since intensity is proportional to the square of the amplitude, the resultant intensity becomes (2a)^2 = 4a^2 = 4I.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In Young's experiment two coherent sources are placed $0.90\ mm$ apart and fringe are observed one metre away. If it produces second dark fringes at a distance of $1\ mm$ from central fringe., the wavelength of monochromatic light is used would be 

  1. $60\ \times 10^{-4}\ cm$
  2. $10\ \times 10^{-4}\ cm$
  3. $10\ \times 10^{-5}\ cm$
  4. $6\ \times 10^{-5}\ cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Formula,

$x=(2n-1)\dfrac{\lambda D}{2d}$

$\Rightarrow \lambda =\dfrac{2xd}{(2n-1)D}$

$=\dfrac{2\times 10^{-3}\times 0.9\times 10^{-3}}{(2\times 2-1)\times 1}$

$=6 \times 10^{-5}cm$