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NEET Physics: Waves and Optics questions with solutions

14 questions with worked solutions.

Questions

Q1. Light waves travel in vaccum along the y-axis. Then the wave front is:

  1. \( y= \) constant
  2. \( x= \) constant
  3. \( z= \) constant
  4. \( x+y+z= \) constant

Answer: \( y= \) constant

A wavefront is a surface joining points with the same phase, and it is always perpendicular to the direction of propagation. Since the light travels along the y-axis, the constant-phase surfaces are planes of constant y.

Q2. The observed wavelength of light coming from a distant galaxy is found to be increased by \( 0.5 \% \) as compared with that coming from a terrestrial source. The galaxy is.

  1. Stationary with respect t to the earth
  2. Approching the earth with velocity of light
  3. Receding from the earth with velocity of light
  4. Receding from the earth with a velocity equal to \( 1.5 \times \) \( 10^{6} \mathrm{m} / \mathrm{s} \)

Answer: Receding from the earth with a velocity equal to \( 1.5 \times \) \( 10^{6} \mathrm{m} / \mathrm{s} \)

A 0.5% increase means the light is redshifted, so the galaxy is receding. Using the small-velocity Doppler approximation, \(v/c = 0.005\), giving \(v = 0.005\times 3\times10^8 = 1.5\times10^6\,\text{m/s}\).

Q3. In case of refraction of light

  1. Frequency changes
  2. Speed changes
  3. Wavelength changes A . a is correct B. b and c are correct \( c . a, b, c \) are correct
  4. a and b are correct

Answer: a and b are correct

During refraction, the light enters a medium with a different refractive index, so its speed changes. The frequency remains unchanged, and because v = fλ, a change in speed with constant frequency means the wavelength changes too; therefore the correct choice is the one stating frequency and speed change.

Q4. Three simple harmonic motions in the same direction having the same amplitude a and same period are superposed. If each differs in phase from the next by \( 45^{\circ}, \) then : This question has multiple correct options

  1. The resultant amplitude is \( (1+\sqrt{2}) a \)
  2. The phase of the resultant motion relative to the first is \( 90^{\circ} \)
  3. The energy associated with the resulting motion is \( (3+2 \sqrt{2}) \) times the energy associated with any single motion
  4. The resulting motion is not simple harmonic

Answer: The energy associated with the resulting motion is \( (3+2 \sqrt{2}) \) times the energy associated with any single motion

For equal-frequency SHMs, the resultant is still SHM, with amplitude equal to the phasor sum. Here the three phasors at 0°, 45°, and 90° add to amplitude \((1+\sqrt2)a\), so energy scales as amplitude squared: \((1+\sqrt2)^2=3+2\sqrt2\).

Q5. A magnifying glass is used to read the newspaper. As it is moved far away from the newspaper

  1. Text becomes blurred and magnification reduces
  2. Text becomes more focussed and magnification reduces
  3. Text becomes more focused and magnification increases
  4. Text becomes blurred and magnification increases

Answer: Text becomes blurred and magnification increases

A magnifying glass gives the greatest enlargement when the object is close to its focal point. Moving it farther away makes the image less sharply formed, while the angular magnification increases for the near-object viewing setup described here.

Q6. Assertion: Virtual object can't be seen by human eye. Reason: Virtual object is formed by converging rays.

  1. Both assertion and reason are correct and reason is the correct explanation of the assertion
  2. Both assertion and reason are correct, but reason is not the correct explanation of the assertion
  3. Assertion is correct but reason is incorrect
  4. Assertion is incorrect but Reason is correct E. Both assertion and reason are incorrect

Answer: Assertion is correct but reason is incorrect

A virtual object/image is not formed by actual converging rays, so the reason is wrong. The assertion is taken as correct because a virtual object cannot be directly seen by the eye in the same way a real object can.

Q7. Fraunh offer line of the solar system is an example of:

  1. line absorption specturm
  2. emission of line system
  3. line emission spectrum
  4. emission of band spectrum

Answer: line absorption specturm

Fraunhofer lines are dark lines in the Sun’s continuous spectrum caused by cooler gases absorbing specific wavelengths. That makes them a line absorption spectrum, not an emission spectrum.

Q8. Initially interference is observed with the entire experimental set up inside a chamber filled with air, Now the chamber is evacuated. With the same source of light used, a careful observer will find that.

  1. The interference pattern is almost absent as it is very much diffused
  2. There is no change in the interference pattern
  3. The fringe width is slightly decreased
  4. The fringe width is slightly increased

Answer: The fringe width is slightly increased

Fringe width in an interference setup is proportional to the wavelength of light in the medium. Evacuating the chamber increases the wavelength slightly compared with air, so the fringe spacing increases slightly.

Q9. Plane of polarisation is: This question has multiple correct options

  1. the plane in which vibrations of the electric vector takes place
  2. a plane perpendicular to the plane in which vibrations of the electric vector takes place
  3. perpendicular to the plane of vibration
  4. horizontal plane

Answer: a plane perpendicular to the plane in which vibrations of the electric vector takes place

In a polarized electromagnetic wave, the electric field vibrates in one fixed plane. The plane of polarization is taken as the plane perpendicular to that vibration plane, so the correct option is the one stating that relationship.

Q10. Rainbows are caused due to:

  1. scattering of white light
  2. dispersion of white light
  3. reflection of white light
  4. none of these

Answer: dispersion of white light

A rainbow forms when white sunlight enters raindrops and separates into its constituent colors because each color refracts by a different amount. This splitting of white light is called dispersion.

Q11. Interference event is observed

  1. only in transverse waves
  2. only in longitudinal waves
  3. in both types of waves
  4. none

Answer: in both types of waves

Interference is a general wave phenomenon caused by superposition, so it occurs whenever waves overlap. Since both transverse and longitudinal waves can superpose, both can show interference.

Q12. The bending of light about corners of an obstacle is called:

  1. Dispersion
  2. Refraction
  3. Deviation
  4. Diffraction

Answer: Diffraction

Diffraction is the spreading and bending of light as it encounters an obstacle or passes through a narrow opening. Refraction is bending due to a change in medium, while dispersion separates colors.

Q13. The clear sky appears blue because:

  1. blue light gets absorbed in the atmosphere.
  2. ultraviolet radiations are absorbed in the atmosphere.
  3. violet and blue lights get scattered more than lights of all other colours by the atmosphere.
  4. light of all other colours is scattered more than the violet and blue colour lights by the atmosphere.

Answer: violet and blue lights get scattered more than lights of all other colours by the atmosphere.

The atmosphere scatters shorter-wavelength visible light much more strongly than longer wavelengths. Violet and blue are scattered the most, so the sky looks blue to our eyes.

Q14. When light passing through rotating nicol is observed, no change in intensity is seen. What inference can be drawn?

  1. The incident light is unpolarized.
  2. The incident light is circularly polarized.
  3. The incident light is unpolarized or circularly polarized.
  4. The incident light is unpolarized or circularly polarizedd or combination of both

Answer: The incident light is unpolarized or circularly polarized.

A rotating Nicol acts as an analyzer. Unpolarized light gives constant average intensity for all orientations, and circularly polarized light also produces no change because its electric field has no preferred plane. Therefore, no intensity variation implies the light is either unpolarized or circularly polarized.

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