Tag: diffraction

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Questions Related to diffraction

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

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

The fringe width beta is given by lambda * D / d. Here, lambda = 5 * 10^-7 m, D = 2 m, d = 10^-3 m. beta = (5 * 10^-7 * 2) / 10^-3 = 10^-3 m = 1.0 mm.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

Each of the four pairs of light waves arrives at a certain point on a screen. The waves have the same wavelength. At the arrival point, their amplitudes and phase differences are :
$2 \mathrm { a } _ { 0 } , 6 \mathrm { a } _ { 0 } $ and $\pi$ rad
$3 \mathrm { a } _ { 0 } , 5 \mathrm { a } _ { 0 } $ and $\pi$ rad
$9 \mathrm { a } _ { 0 } , 7 \mathrm { a } _ { 0 } $ and $3\pi$ rad
$2 \mathrm { a } _ { 0 } , 2\mathrm { a } _ { 0 } $ and $0$
The pair/s which has greatest intensity is /are :

  1. $I$
  2. $II$
  3. $II, III$
  4. $I, IV$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Intensity I is proportional to the square of the resultant amplitude. Resultant amplitude R = sqrt(A1^2 + A2^2 + 2*A1*A2*cos(phi)). For I and IV, the phase differences are pi and 0 respectively, leading to high resultant amplitudes.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In Young's double slit. experiment, distance between two sources is 0.1 mm. The distance of screen from the sources is 20 cm. Wavelength of light used is 5460 k Then angular position of first dark fringe is 

  1. $0.20 ^ { \circ }$
  2. $0.32 ^ { \circ }$
  3. $0.08 ^ { \circ }$
  4. $0.16 ^ { \circ }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The path difference for the first dark fringe is lambda/2. The angular position theta is given by sin(theta) = path_diff / d = (lambda/2) / d. With lambda = 5460 * 10^-10 m and d = 10^-4 m, sin(theta) = 2.73 * 10^-3. theta is approximately 0.16 degrees.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

The waves of $600 \mu m$ wave length are incident normally on a slit of $1.2\ mm$ width. The value of diffraction angle corresponding to the first minima will be (in radian):

  1. $\dfrac{\pi}{2}$
  2. $\dfrac{\pi}{6}$
  3. $\dfrac{\pi}{5}$
  4. $\dfrac{\pi}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given,

$\lambda=600\mu m$
$d=1.2mm$
condition for minima 
$dsin\theta=m\lambda$
for $m=1$
$sin\theta=\dfrac{m\lambda}{d}=\dfrac{\lambda}{d}$

$sin\theta=\dfrac{600\times 10^{-6}}{1.2\times 10^{-3}}=0.5$

$\theta=sin^{-1}(0.5)=\dfrac{\pi}{6}$
The correct option is B.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In the case of interference, The maximum and minimum intensities are in the ratio $16:9$. Then

  1. The maximum and minimum amplitude will be in the ratio 9:5

  2. The intensities of the individual waves will be in the ratio 4:3.

  3. The amplitudes of the individual waves will be in the ratio 4:1

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

Given I_max / I_min = 16 / 9, the ratio of amplitudes a_1 / a_2 can be found since I proportional to a^2. Taking square roots gives (a_1 + a_2) / (a_1 - a_2) = 4 / 3, which solves to a_1 / a_2 = 7:1. Since the options given do not match this correct calculation, the right choice is none of the above.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In young's double slits experiments , the distance between the slits is 1 mm and that between slit and screen is 1 meter and 10th fringe is 5 mm away from the central bright fringe, then wavelength of light used will be

  1. $ 8000 A^0 $
  2. $ 7000 A^0 $
  3. $ 6000 A^0 $
  4. $ 5000 A^0 $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the fringe width formula x_n = n * lambda * D / d, we substitute x_10 = 5 mm, n = 10, D = 1 m, and d = 1 mm. Solving for lambda gives lambda = (5 * 10^-3 * 10^-3) / (10 * 1) = 5 * 10^-7 m, which equals 5000 angstroms.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

In young's double slit experiment, when two light waves form third minimum intensity, they have 

  1. phase difference of 3$\pi$
  2. phase difference of $\dfrac { 5 \pi } { 2 }$
  3. path difference of 3$\lambda$
  4. path difference of $\dfrac { 5 \lambda } { 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The m-th minimum in a double slit interference pattern occurs at a path difference of delta = (2m - 1) * lambda / 2. For the third minimum (m = 3), the path difference is (2(3) - 1) * lambda / 2 = 5 * lambda / 2.

Multiple choice difference between interference and diffraction explaining wave phenomena diffraction

For constructive interference to take place between two monochromatic light waves of wavelength $  \lambda  $ , the path difference should be

  1. $
    (2 n-1) \frac{\lambda}{4}
    $
  2. $
    (2 n-1) \frac{\lambda}{2}
    $
  3. $
    n \lambda
    $
  4. $
    (2 n+1) \frac{\lambda}{4}
    $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Constructive interference occurs when the path difference is an integer multiple of the wavelength, i.e., n * lambda.