Wave Optics Quiz-20

Dear Readers,

As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .

Q1.  . At two points 𝑃 and 𝑄 on screen in Young's double slit experiment. Waves from slits 𝑆1 and 𝑆 have a path difference of 0 and πœ†/4 respectively. The ratio of intensities at P and Q will be
  •   3 ∶ 2
  •   2 ∶ 1
  • √2 ∶ 1
  •   4 ∶ 1
Let 𝐼0 is intensity of light emitted from the source,
then Resultant intensity
 πΌ = 4𝐼0 cos2 πœ™/2 
 πΌ1 = 4𝐼0 cos2 πœ™/2 = 4𝐼0 
Now, ∆π‘₯ = πœ†/4 
 Ο• =2πœ‹/πœ†× ∆π‘₯ =2πœ‹/πœ†×πœ†/4 
 Ο• = πœ‹/2 And 
 πΌ2 = 4𝐼0 cos2 πœ‹/4 = 2𝐼0
  𝐼1:𝐼2 = 2 ∶ 1

Q2. Wave nature of light follows because
  •   Light rays travel in a straight line
  •   Light exhibits the phenomena of reflection and refraction
  •   Light exhibits the phenomena of interference
  •   Light causes the phenomena of photoelectric effect
Interference is explained by wave nature of light

Q3.   The angular width of the central maximum of the diffraction pattern in a single slit (of width ′π‘Ž′) experiment, with πœ† as the wavelength of light is
  •   3πœ†/2π‘Ž
  •   πœ†/2π‘Ž
  •   2πœ†/π‘Ž
  •   πœ†/π‘Ž

Q4.  In Young’s double slit experiment, if monochromatic light is replaced by white light
  •   All bright fringes become white
  •   All bright fringes have colours between violet and red
  •   Only the central fringe is white, all other fringes are coloured
  •   No fringes are observed
In Young’s double slit experiment, if white light is used in place of monochromatic light, then the central fringe is white and some coloured fringes around the central fringe are formed 

Since 𝛽red > 𝛽violet etc., the bright fringe of violet colour forms first and that of the red forms later It may be noted that, the inner edge of the dark fringe is red, while the outer edge is violet. Similarly, the inner edge of the bright fringe is violet and the outer edge is red

Q5. In Young’s experiment, the distance between slits is 0.28 π‘šπ‘š and distance between slits and screen is 1.4 π‘š. Distance between central bright fringe and third bright fringe is 0.9 π‘π‘š. What is the wavelength of used light
  •   5000 β„«
  •   6000 β„«
  •   7000 β„«
  •   9000 β„«
  Position of 3π‘Ÿπ‘‘ bright fringe π‘₯3 = 3π·πœ†/𝑑 
⇒ πœ† = π‘₯3𝑑/3𝐷 = (0.9 × 10-2) × (0.28 × 10-3)/3 × 1.4 = 6000β„« .

Q6.  In a Young’s double slit experiment(slit distance d) monochromatic light of wavelength πœ† is used and the fringe pattern observed at a distance 𝐿 from the slits. The angular position of the bright fringes are
  •   sin-1 ( π‘›πœ†/𝑑 )
  •   sin−1 ( (𝑛 + 1/2 )πœ†/𝑑 )
  • sin-1 ( π‘›πœ†/L)
  •   sin-1 ( [(𝑛 + 1/2 )πœ†]/ 𝐿 )
For constructive interference Path difference ∆ = 𝑑 sinΞΈ = π‘›πœ† ⇒ πœƒ = sin-1[ π‘›πœ†/𝑑 ]

Q7. 𝑛 coherent source of intensity 𝐼0 are superimposed at a point, the intensity of the point is  

  •   𝑛𝐼0
  •   𝐼0/𝑛
  •   𝑛2𝐼0
  •   none of these

Q8. When a compact disc is illuminated by small source of white light, coloured bands are observed. This is due to
  •   Dispersion
  • Diffraction
  •   Interference
  •   Reflection
The data which represents the music is stored on the compact disc in the form of very small pits arranged in a tightly wound spiral track in silvery surface. The distance between two neighbouring track is 1.6 micrometre. Which is only several times the wavelength of visible light, this small spacing is responsible for the wonderful colours reflected by a CD which works as a diffraction grating. Hence, diffraction is responsible for coloured bands.

Q9. . If Young’s double slit experiment, is performed in water

  •   The fringe width will decrease
  •   The fringe width will increase
  •   The fringe width will remain unchanged
  •   There will be no fringe
As we know
 Ξ² = 𝐷/𝑑 πœ† 
 Ξ» ∝1/ΞΌ 
 From Eqs. (i) and (ii), 
 Ξ² ∝ Ξ» ∝1/ΞΌ
 Ξ² ∝1/ΞΌ 
 The refractive index of water is greater than air, therefore fringe width will decrease

Q10.  For the constructive interference the path difference between the two interfering waves must be equal to  


  •   (2𝑛 + 1)πœ†
  •   2π‘›πœ‹
  •   π‘›πœ†
  •   (2𝑛 + 1)πœ†/2
Phase difference, 
 ∆Ο• = 2πœ‹/πœ† ∆π‘₯ 
 In a constructive interference,
 ∆Ο• = 2π‘›πœ‹
 Where 𝑛 = 0,1,2,3,…… 
 ∴ 2π‘›πœ‹ = 2πœ‹/πœ† ∆π‘₯ 
Or ∆π‘₯ = π‘›πœ†

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