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  WAVE OPTICS Quiz-11

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.  In a Young’ s double slit experiment, the intensity at a point where the path difference is πœ†/6 where(Ξ» is wavelength of the light) is 𝐼. I f 𝐼0denotes the maximum intensity, then 𝐼/𝐼o is equal to
  •   1/2
  •   √3/2
  •   1/√2
  •   3/4
Solution
Ο• = πœ†/6 = 360°/6 = 60° 𝐼 = 𝐼0 cos2 πœƒ = 𝐼0 cos2 60° =3/4× πΌ0 𝐼/𝐼0 = 3/4

Q2. A beam with wavelength πœ† falls on a stack of partially reflecting planes with separation 𝑑. The angle πœƒ that the beam should make with the planes so that the beams reflected from successive planes may interfere constructively is (where 𝑛 = 1,2,……) 

 
  •   sin-1 (π‘›πœ†/𝑑 )
  •   tan-1( π‘›πœ†/𝑑 )
  •   sin-1 ( π‘›πœ†/2𝑑 )
  •   cos-1 ( π‘›πœ†/2𝑑 )
Solution
Path difference = 2𝑑 sinπœƒ ∴ For constructive interference 2𝑑 sinπœƒ = π‘›πœ† ⇒ πœƒ = sin-1 ( π‘›πœ†/2𝑑 ) 




Q3.   S1 And S2 are two coherent sources. The intensity of both sources are same. If the intensity at the point of maxima is 4 Wm-2, the intensity of each source is
  •   1 Wm-2
  •   2 Wm-2
  •   3 Wm-2
  •   4 Wm-2
Solution
At a point of maxima ∴ 𝐼max = 4𝐼0 = 4 Wm-2 ∴ 𝐼0 = 1 Wm-2

Q4.  A single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength 6500β„«?
  •   4400 β„«
  •   4100 β„«
  •   4642.8 β„«
  •   9100 β„«
Solution
π‘₯ = (2𝑛 + 1)πœ†π·/2π‘Ž For red light π‘₯ = (4+1)𝐷/2π‘Ž× 6500 For unknown wavelength of light, π‘₯ = (6 + 1)𝐷/2π‘Ž × πœ† Accordingly ∴ 5 ×6500 = 7 × πœ† ⇒ πœ† = 5/7 × 6500 = 4642.8 β„«


Q5. In diffraction from a single slit, the angular width of the central maxima does not depend on
  •   πœ† of light used
  •   Width of slit
  •   Distance of slits from screen
  •   Ratio of πœ† and slit width
Solution
  For a slit of width π‘Ž, light of wavelength πœ†, when light falls on the slit, the diffraction patterns so obtained as 

 The first diffraction minimum occurs at the angles given by sinΞΈ = πœ†/π‘Ž 
From the equation, it is clear that width of the central diffraction maximum is inversely proportional to the width of the slit. On increasing the width sizeπ‘Ž, the angle ΞΈ at which the intensity first becomes zero decreases, resulting in a narrower central band and if the slit width is made smaller, the angle ΞΈ increases, giving a wider central band .

Q6.  The Young’s double slit experiment is performed with blue and with green light of wavelength 4360 β„« and 5460 β„« respectively. If π‘₯ is the distance of 4th maxima from the central one, then
  •   π‘₯(blue) = π‘₯(green)
  •   π‘₯(blue) > π‘₯(green)
  • π‘₯(blue) < π‘₯(green)
  •   π‘₯(blue)/π‘₯(green) = 5400/4360
Solution
Distance of π‘›π‘‘β„Žmaxima, π‘₯ = π‘›πœ† 𝐷/𝑑 ∝ πœ† As πœ†π‘ < πœ†π‘” ∴ π‘₯blue < π‘₯green


Q7. TV waves have a wavelength range of 1-10 π‘šπ‘’π‘‘π‘’π‘Ÿ. Their frequency range in 𝑀𝐻𝑧 is  

 
  •   30-300
  •   3-30
  •   300-3000
  •   3-3000
Solution

𝑣 = 𝐢/πœ† ⇒ 𝑣1 = (3 × 108)/1 = 3 ×108𝐻𝑧 = 300 𝑀𝐻𝑧 and 𝑣2 = 3×108/10 = 3 × 107𝐻𝑧 = 30𝑀𝐻𝑧


Q8. . πœ†π‘Ž and πœ†π‘š are the wavelength of a beam of light in air and medium respectively. If ΞΈ is the polarising angle, the correct relation between πœ†π‘Ž,πœ†π‘š and ΞΈ is
  •   πœ†π‘Ž = πœ†π‘š tan2 ΞΈ
  •   πœ†π‘š = πœ†π‘Ž tan2 ΞΈ
  •   πœ†π‘Ž = πœ†π‘š cotΞΈ
  •   πœ†π‘š = πœ†π‘Ž cotΞΈ
We know, πœ†π‘š = π‘Ž/πœ‡ and πœ‡ = tanΞΈ ∴ πœ†π‘š = πœ†π‘Ž/tanΞΈ = πœ†π‘Ž cotΞΈ


Q9. Which of the following cannot be polarized?

  •   Ultraviolet rays
  •   Ultrasonic waves
  •   X-rays
  •   Radiowaves
Solution
Ultrasonic waves cannot be polarized.

Q10.  In a biprism experiment, 5th dark fringe is obtained at a point. If a thin transparent film is placed in the path of one of waves, then 7th bright fringes is obtained at the same point. The thickness of the film in terms of wavelength 𝑙 and refractive index πœ‡ will be  

 

  •   1.5πœ† /(πœ‡ − 1)
  •   1.5(πœ‡ − 1)πœ†
  •   2.5(πœ‡ − 1)πœ†
  •   2.5πœ†/(πœ‡ − 1)
Solution
For 5th dark fringe, π‘₯1 = (2𝑛 − 1) πœ†/'2 𝐷/ 𝑑 = 9πœ†π· /2𝑑
 For 7th bright fringe, π‘₯2 = π‘›πœ† 𝐷/𝑑 = 7πœ†π·/𝑑
 π‘₯2 − π‘₯1 = (πœ‡ − 1)𝑑 𝐷/𝑑 
 πœ†π·/𝑑 [7 − 9/2 ] = (πœ‡ − 1)𝑑 𝐷/𝑑
 π‘‘ = 2.5πœ†/(πœ‡ − 1)

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