Way Optics Quiz-14

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.  A beam of plane polarized light falls normally on a polarizer of cross sectional area 3 × 10-4π‘š2. Flux of energy of incident ray in 10-3π‘Š. The polarizer rotates with an angular frequency of 31.4 π‘Ÿπ‘Žπ‘‘/𝑠. The energy of light passing through the polarizer per revolution will be
  •   10-4π½π‘œπ‘’π‘™π‘’
  •   10-3π½π‘œπ‘’π‘™π‘’
  •   10-2π½π‘œπ‘’π‘™π‘’
  •   10-1π½π‘œπ‘’π‘™π‘’
Using Malus law, 𝐼 = 𝐼o cos2 πœƒ 
As here polarizer is rotating, 𝑖.𝑒., all the values of πœƒ are possible 
 πΌπ‘Žπ‘£ = 1/2πœ‹ ∫2πœ‹o𝐼 π‘‘πœƒ = 1/2πœ‹ ∫2πœ‹o 𝐼o cos2 πœƒ π‘‘πœƒ 
 On integration we get 𝐼av = 𝐼o/2 
 Where 𝐼o = Energy /Area×Time = 𝑝/𝐴 = 10-3 /3×10-4 = 10 /3 π‘Šπ‘Žπ‘‘π‘‘ /π‘š2 
 ∴ 𝐼av = 1/2 × 10/3 = 5/3 π‘Šπ‘Žπ‘‘π‘‘ and Time period 𝑇 = 2πœ‹/πœ” = 2×3.14 /31.4 = 1/5𝑠 
∴ Energy of light passing through the polarizer per revolution = 𝐼av × Area × π‘‡ = 5/3 × 3 × 10-4 × 1/5 = 10-4𝐽

Q2. In a wave, the path difference corresponding to a phase difference of πœ™ is
  •   πœ‹/2πœ† πœ™
  •   πœ‹ /πœ† πœ™
  •   πœ†/2πœ‹ πœ™
  •   πœ†/ πœ‹ πœ™
For 2πœ‹ phase difference → Path difference is πœ† ∴ For πœ™ phase difference → Path difference is πœ† 2πœ‹ × πœ™

Q3.   Which one of the following property of light does not support wave theory of light?
  •   Light obeys laws of reflection and refraction
  •   Light waves get polarized
  •   Light shows photoelectric effect
  •   Light shows interference
Photoelectric effect does not support the wave theory of light

Q4.  Light waves travel in vacuum along the 𝑦– axis. Which of the following may represent the wavefront?
  •   𝑦 = constant
  •   π‘₯ = constant
  •   z = constant
  •   x+y+z= constant
As velocity of light is perpendicular to the wavefront, and light is travelling in vacuum along the𝑦 − axis, therefore, the wavefront is represented by𝑦 = constant.

Q5. In the given arrangement, 𝑆1 and 𝑆2 are coherent sources (shown in figure). The point P is a point of 

  •   Bright fringe
  •   Dark fringe
  •   Either dark or light
  •   None of the above
  Bright fringe .

Q6.  A wavefront presents one, two and three HPZ at points 𝐴,𝐡 and 𝐢 respectively. If the ratio of consecutive amplitudes of HPZ is 4 : 3, then the ratio of resultant intensities at these point will be
  •   169 : 16 : 256
  •   256 : 16 : 169
  • 256 : 16 : 196
  •   256 : 196 : 16
𝐼A = 𝑅12 
𝐼B = (𝑅1 − 𝑅2)2 = 𝑅12 (1 − 𝑅2/ 𝑅1)2 = 𝑅12 (1 − 3/4 )2 
 = 𝑅12/16
 πΌC = (𝑅1 − 𝑅2 + 𝑅3)2 = 𝑅12 (1 − 𝑅2/ 𝑅1 + 𝑅3/ 𝑅1 )2 
 = 𝑅12 (1 − 𝑅2 /𝑅1 + 𝑅3 /𝑅2 × π‘…2 / 𝑅1)2 = 𝑅12 (1 − 3/4 + 3/4 × 3/4 )2 
 = ( 13/16 )2 𝑅12 = 169/256 𝑅12 
∴ 𝐼A:𝐼B:𝐼C = 𝑅12: 𝑅12/ 16 : 169/256 𝑅12 = 256:16:169

Q7. Irreducible phase difference in any wave of 5000 β„« from a source of light is
  •   πœ‹
  •   12πœ‹
  •   12πœ‹ × 106
  •   πœ‹ × 106
Solution   πœ‹

Q8. In Young’s double slit experiment, the slits are 3 mm apart. The wavelength of light used is 5000 β„« and the distance between the slits and the screen is 90 cm. The fringe width in 9 (mm) is
  •   1.5
  • 0.015
  •   2.0
  •   0.15
Let πœ† be wavelength of monochromatic light, used to illuminate the slit S, and d be the distance between coherent sources, then width of slits is given by 
 π‘Š = π·πœ†/𝑑 
 When D is distance between screen and source. 

 Given, 𝑑 = 3 mm,πœ† = 5000 β„« = 5 × 10-7 m = 5 ×10-4 mm 𝐷 = 90 cm = 900 mm ∴ π‘Š =(5 × 10-4 × 900)/3 = 15 × 10-2 mm = 0.15 mm

Q9.. The figure here gives the electric field of an EM wave at a certain point and a certain instant. The wave is transporting energy in the negative 𝑧 direction. What is the direction of the magnetic field of the wave at that point and instant

  •   Towards + 𝑋 direction
  •   Towards −𝑋 direction
  •   Towards + 𝑍 direction
  •   Towards −𝑍 direction
The direction of EM wave is given by the direction of 𝐸 ⃗ × π΅ ⃗

Q10.  Two polaroids are kept crossed to each other. Now one of them us rotated through an angle of 45°. The percentage of incident light now transmitted through the system is
  •   15%
  •   25%
  •  50%
  • 60%
If 𝐼0 is intensity of unpolarized light, then intensity of polarized light from 1st Polaroid=𝐼o/ 2. 
 On rotating through 45°, intensity of light from 2nd Polaroid, 
𝐼 = (𝐼o /2)(cos45°)2 = 𝐼0/ 2 ( 1/ √2 )2 = 𝐼o/ 4
 = 25% 𝐼o

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