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Karnataka PUC II Dual Nature of Radiation and Matter — practice questions

Explore 98 free Dual Nature of Radiation and Matter questions in Karnataka PUC II Physics. Open each question or exercise to read its content, reference answers and available explanation.

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In Young's double-slit experiment, the fringe width $w$ is given by:Read the answer and worked solution →In Young's experiment, if the entire setup is immersed in a medium of refractive index $n$, the fringe width:Read the answer and worked solution →Two slits 0.5 mm apart are illuminated with light of 600 nm. The fringes are observed on a screen 1 m away. The fringe width is:Read the answer and worked solution →Brewster's angle for light going from air (n=1) to glass (n=1.5) is approximately:Read the answer and worked solution →A telescope objective has diameter 2.5 m. Its angular resolution at $\lambda = 5500$ Å is approximately:Read the answer and worked solution →The optical path of a 3 cm ray in flint glass (n = 1.6) equals that of an unknown medium (n = 1.25) over distance $x$. Then $x$ is:Read the answer and worked solution →Light enters a glass slab (n = 1.5) at angle i = 30°. The angle of refraction r inside glass is approximately:Read the answer and worked solution →For two equal-amplitude coherent sources, the maximum and minimum intensities are:Read the answer and worked solution →Snell's law states that the ratio sin(i)/sin(r) for two media is:Read the answer and worked solution →The wavefront from a point source is:Read the answer and worked solution →Young's double-slit experiment demonstrates which phenomenon?Read the answer and worked solution →Two waves with the same frequency and constant phase difference are called:Read the answer and worked solution →In Young's double-slit experiment, the bright fringe condition is path difference equals:Read the answer and worked solution →Polarization of light is possible because light is:Read the answer and worked solution →In Young's experiment, fringe width β equals:Read the answer and worked solution →In Young's experiment with d = 1 mm, D = 1 m, light wavelength 600 nm, fringe width is:Read the answer and worked solution →When light goes from a less dense to denser medium (e.g., air to water), it bends:Read the answer and worked solution →Photoelectric effect was explained by:Read the answer and worked solution →Photon energy:Read the answer and worked solution →de Broglie wavelength λ associated with a particle of momentum p:Read the answer and worked solution →In photoelectric effect, intensity of light affects:Read the answer and worked solution →Einstein's photoelectric equation:Read the answer and worked solution →Threshold frequency ν₀ for photoemission:Read the answer and worked solution →de Broglie wavelength of electron accelerated through V volts:Read the answer and worked solution →For very slow electron (low momentum), de Broglie wavelength is:Read the answer and worked solution →Davisson-Germer experiment demonstrated:Read the answer and worked solution →For photon, momentum p = E/c =Read the answer and worked solution →A metal has work function 2.0 eV. Light of wavelength 400 nm (3.1 eV) shines on it. Max KE of photoelectrons:Read the answer and worked solution →Threshold wavelength for a metal with work function 3 eV:Read the answer and worked solution →de Broglie wavelength of a 1 kg ball moving at 1 m/s:Read the answer and worked solution →

Worked practice questions

In Young's double-slit experiment, the fringe width $w$ is given by:

  1. $w = \lambda d / D$
  2. $w = \lambda D / d$
  3. $w = D / (\lambda d)$
  4. $w = (\lambda + D)/d$
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Correct answer: B. $w = \lambda D / d$

Fringe width (separation between adjacent maxima or minima) $w = \lambda D/d$, where $D$ = screen-slit distance, $d$ = slit separation, $\lambda$ = wavelength. Bigger $\lambda$ or $D$, or smaller $d$ ⇒ wider fringes.

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In Young's experiment, if the entire setup is immersed in a medium of refractive index $n$, the fringe width:

  1. Stays the same
  2. Increases by factor $n$
  3. Decreases by factor $n$ (since $\lambda$ shortens)
  4. Becomes zero
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Correct answer: C. Decreases by factor $n$ (since $\lambda$ shortens)

Wavelength in medium $\lambda' = \lambda/n$, so fringe width $w' = \lambda' D/d = w/n$. **Hence immersion in water (n=1.33) shrinks fringes by ~25%.**

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Two slits 0.5 mm apart are illuminated with light of 600 nm. The fringes are observed on a screen 1 m away. The fringe width is:

  1. 0.6 mm
  2. 1.2 mm
  3. 1.5 mm
  4. 3 mm
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Correct answer: B. 1.2 mm

We have $w = \lambda D/d$. That is $w = (600 \times 10^{-9})(1)/(0.5 \times 10^{-3})$. Then $w = 1.2 \times 10^{-3}$. So m = 1.2 mm.

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