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.  If a source of light is moving away from a stationary observer, then the frequency of light wave appears to change because of
  •   Doppler’s effect
  •   Interference
  •   Diffraction
  •   None of these
According to Doppler’s effect, wherever there is a relative motion between source and observer, the frequency observed is different from that given out by source

Q2. . In a Young’s double slit experiment, the fringe width will remain same, if (𝐷 = distance between screen and plane of slits, 𝑑 = separation between two slits and πœ† = wavelength of light used)
  •   Both πœ† and 𝐷 are doubled
  •   Both 𝑑 and 𝐷 are doubled
  •   𝐷 is doubled but 𝑑 is halved
  •   πœ† is doubled but 𝑑 is halved
𝛽 = πœ†π·/𝑑

Q3.   Two parallel slits 0.6 π‘šπ‘š apart are illuminated by light source of wavelength 6000 β„«. The distance between two consecutive dark fringes on a screen 1 π‘š away from the slits is
  •   1 π‘šπ‘š
  •   0.01 mm
  •   0.1 π‘š
  •   10 π‘š
Distance between two consecutive Dark fringes = πœ†π·/𝑑 =6000×10-10×1) /0.6×10-3 = 1 ×10−3π‘š = 1π‘šπ‘š

Q4.  A single slit Fraunhoffer 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 6500 β„«
  •   4400 β„«
  •   4100 β„«
  •   4642.8 β„«
  •   9100 β„«
π‘₯ = (2𝑛 + 1)πœ†π·/ 2π‘Ž
 For red light, π‘₯ = (4+1)𝐷 /2π‘Ž× 6500β„«
 For other light, π‘₯ =(6+1)𝐷 /2π‘Ž× πœ†β„« 𝑋 is same for each 
∴ 5 ×6500 = 7 × πœ† ⇒ πœ† =5/7 × 6500 = 4642.8β„«

Q5. Consider the following statements A and B and identify the correct answer. A. Fresnel’s diffraction pattern occurs when the source of light or the screen on which the diffraction pattern is seen or when both are at finite distance from the aperture. B. Diffracted light can be used to estimate the helical structure of nucleic acids
  •   A and B are true
  •   A and B are false
  •   A is true but B is false
  •   A is false but B is true
  A is true but B is false .

Q6.  In a Young’s double slit experiment, the separation between the two slits is 0.9 mm and the fringes are observed 1 m away. If it produces the second dark fringes at a distance of 1 mm from the central fringe, the wavelength of the monochromatic source of light used is
  •   450 nm
  •   400 nm
  • 5002 nm
  •   600 nm
Given, spacing between second dark fringe and central fringe = 𝛽 + 𝛽/ 2 Or 3𝛽/2 = 1 mm or 𝛽 = 2/3 × 1 mm πœ†π· /𝑑 = 2/ 3 mm 
 ∴ πœ† = 2 /3 × 10-3 × 0.9 × 10-3 /𝟏
 ∴ πœ† = 0.6 ×10-6 m
 ∴ πœ† = 600 × 10-9 m = 600 m

Q7. In a Young’s experiment, two coherent sources are placed 0.90 mm apart and the finges are observed one metre away. If it produces the second dark fringe at a distance of 1 mm from the central fringe, the wavelength of monochromatic light used would be
  •   60 × 10-4cm
  •   10 × 10-4cm
  •   10 × 10-5cm
  •   6 ×10-5cm

Distance of 𝑛th dark fringe from central fringe π‘₯𝑛 = (2𝑛 − 1)πœ†π· /2𝑑 
 ∴ π‘₯2 = (2 × 2 − 1)πœ†π· /2𝑑 = 3πœ†π·/ 2𝑑 
 ⇒ 1 × 10-3 = (3πœ† × 1)/( 2 × 0.9 × 10-3) ⇒ πœ† = 6 × 10-5cm

Q8. Through which character we can distinguish the light waves from sound waves
  •   Interference
  • Refraction
  •   Polarization
  •   Reflection
Polarization is not shown by sound waves

Q9. In Young’s double slit experiment, the separation between the slit and the screen increases. The fringe width

  •   Increases
  •   Decreases
  •   Remains unchanged
  •   None of these
𝛽 = πœ†π·/𝑑

Q10.  The π‘˜ line of singly ionized calcium has a wavelength of 393.3 π‘›π‘š as measured on earth. In the spectrum of one of the observed galaxies, this spectral line is located at 401.8 π‘›π‘š. The speed with which the galaxy is moving away from us, will be
  •   6480 π‘˜π‘š/𝑠
  •   3240 π‘˜π‘š/𝑠
  •   4240 π‘˜π‘š/𝑠
  • None of these
∆πœ† πœ† = 𝑣/𝑐 ⇒ (401.8 − 393.3) /393.3 = 𝑣 /(3 × 108) ⇒ 𝑣 = 6.48 × 106π‘š/𝑠 = 6480π‘˜π‘š/𝑠

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