Tag: diffraction

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Questions Related to diffraction

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

The yellow light source in Young's double- slit experiment is replaced by a monochromatic red light source of same intensity. Then the fringe width of the interference pattern in comparison with that of the previous pattern will 

  1. increase

  2. decrease

  3. remain unchanged

  4. vanish
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Fringe width beta = lambda * D / d. Since the wavelength of red light is longer than that of yellow light, the fringe width increases.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In a Young's double slit experiment, $d = 1\ mm,$ $\lambda = 6000 \overset {0}{A}$ & $D = 1\ m.$ The slits produce same intensity on the screen. The minimum distance between two points on the screen having $75\%$ intensity of the maximum intensity is:

  1. $0.45\ mm$
  2. $0.40\ mm$
  3. $0.30\ mm$
  4. $0.20\ mm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The intensity I at a point with phase difference phi is I = I_max * cos^2(phi/2). For I = 0.75 * I_max, cos^2(phi/2) = 3/4, so phi/2 = pi/6 or 5pi/6. The path difference is delta = (lambda/2pi) * phi, and y = (D/d) * delta. Calculating the positions for both phase values gives the distance between them as 0.40 mm.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

Direction:
The question has a paragraph followed by two statements, Statement-1 and Statement-2. Of the given four alternatives after the statements, choose the one that describes the statements.
A thin air film is formed by putting the convex surface of a plane-convex lens over a plane glass plate. 
With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film. 
Statement -1 : When light reflects from the air-glass plate interface, the reflected wave suffers a wave change of $\pi $
Statement-2: The centre of the interference pattern is dark

  1. Statement-1 is true, Statement-2 is false.

  2. Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of Statement-1

  3. Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of Statement-1

  4. Statement-1 is false, Statement-2 is true
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As the center of the pattern is dark, we know that the phase difference is $(2n+1)\pi$. Furthermore, we also know that for the center of the interface pattern, there is no path difference. So the difference can only be due to reflection which is $\pi$. Therefore, both the assertion and the reason are correct and the reason is explanatory,

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

The distance of n bright fringe to the $(n+1)^{th}$ dark fringe in Young's experiments is equal to : 

  1. $\frac{n\lambda D}{d}$
  2. $\frac{n\lambda D}{2d}$
  3. $\frac{\lambda D}{2d}$
  4. $\frac{\lambda D}{d}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The distance from the nth bright fringe to the (n+1)th dark fringe corresponds to half of a fringe width. Since the fringe width is beta = lambda * D / d, half of this width is lambda * D / (2d).

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In a young's bouble slit experiment ,$d=1mm,\lambda  = 6000\mathop A\limits^0 $ and $D=1m$(where d,$\lambda$ and D have unit meaning). Each of slit individually produces same intensity on the screen. The minimum distance between two points  on the screen having $75$% intensity of the maximum intensity is:

  1. $0.45mm$
  2. $0.40mm$
  3. $0.30mm$
  4. $0.20mm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

 In Young's experiment , the separation between 5th maxima and 3rd minima is how many times as that of fringr width?

  1. 5 time

  2. 3 times

  3. 2 times

  4. 2.5times
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
For interference pattern, in double slit experiment maxima will be found at a distance y from the central maximum,
$y=n\lambda D/d$
$D=$distance between slit and screen
$d=$ separation of the slits
For $n=5$,
$y=5\lambda D/d$
For minima
$y=(2n+1)\lambda D/2d$
For $n=3$,
$y=7\lambda D/d$
Difference between maxima and minima$=7\lambda D/d-5\lambda D/d=2\lambda D/d$
Now, fridge width is given by $x=\lambda D/d$
Thus, the given difference is $2$ times the fringe width.
Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

Monochromatic light of wavelength 580 mm incident on a slit of width 0.30 mm. The screen 2 m from the slit. the width of the center maximum is 

  1. $3.35\times 10^{-3}m$
  2. $2.25\times 10^{-3}m$
  3. $6.20\times 10^{-3}m$
  4. $7.7\times 10^{-3}m$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$d=580nm=580\times {10}^{-9}m$

$d=0.3mm=0.0003m$
width of the cental maxima $=\cfrac{2\lambda D}{d}$
$=\cfrac{2\times 580\times {10}^{-9}\times 2}{0.0003}$
$=0.0077m$

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

Monochromatic green light of wavelength 5 $\times 10^{-7}$ m illuminates a pair of slits 1 mm apart. The separation of bright lines in the interference pattern formed on a screen 2 m away is :

  1. 0.25 mm

  2. 0.1 mm

  3. 1.0 mm

  4. 0.01 mm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,


$\lambda=5\times 10^{-7}m$


$d=1mm=0.001m$

$D=2m$

The separation between two bright fringe is called fringe width.

$B=\dfrac{D\lambda}{d}$

$B=\dfrac{2\times 5\times 10^{-7}}{10^{-3}}$

$B=1mm$

The correct option is C.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

A metal plate is placed 2 m away from a monochromatic light source of 1 mW power. Assuming that an electron in metal collects its energy from a circular area of the plate as large as 10 atomic diameters (${10^{ - 9}}\;m$) in radius, calculate how long it will take for such a 'target' to 'soak off' % eV of energy for its emission from the metal?

  1. 1.5 hr

  2. 2.5 hr

  3. 3.5 hr

  4. 4.5 hr
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The power is spread over a sphere of area 4π(2)^2 = 16π m^2 at distance 2m. The electron collects from area π(10^-9)^2 m^2, so the fraction is 10^-18/16 = 6.25×10^-20. Power on target = 1 mW × 6.25×10^-20 = 6.25×10^-23 W. For 5 eV (5 × 1.6×10^-19 J), time = 8×10^-19 / 6.25×10^-23 ≈ 12800 s ≈ 3.5 hr. The question likely meant 5 eV (typo: % eV).

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

To make the central fringe at the centre O, a mica sheet of refractive index $1.5$ is introduced. Choose the correct statements (s).

  1. The thickness of sheet is $2\left( {\sqrt 2 - 1} \right)d$ in front of ${S _1}$
  2. The thickness of sheet is $\left( {\sqrt 2 - 1} \right)d$ in front of ${S _2}$
  3. The thickness of sheet is $2\sqrt 2 d$ in front of ${S _1}$
  4. The thickness of sheet is $\left( {2\sqrt 2 - 1} \right)d$ in front of ${S _1}$
Reveal answer Fill a bubble to check yourself
A Correct answer