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ISC Class 12 Dual Nature of Radiation and Matter — practice questions

Explore 50 free Dual Nature of Radiation and Matter questions in ISC Class 12 Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Dual Nature of Radiation and Matter questions and exercises with answers

Showing questions 1–30 of 50.

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 →Knock out the photoelectric mythsRead the answer and worked solution →Who proposed that matter also has a wave nature?Read the answer and worked solution →In a photoelectric experiment, the minimum negative potential applied to the collector plate at which the photocurrent just drops to zero is called theRead the answer and worked solution →Name the physicistRead the answer and worked solution →True or false about photoemissionRead the answer and worked solution →Dual nature bingoRead the answer and worked solution →Complete the photoelectric ruleRead the answer and worked solution →NEET · Physics · Photoelectric — Spot the ErrorRead the answer and worked solution →Work function is usually measured in:Read the answer and worked solution →Photoelectric emission was discovered in 1887 by:Read the answer and worked solution →Below the threshold frequency, photoelectric emission:Read the answer and worked solution →Who explained the photoelectric effect in 1905?Read the answer and worked solution →In Einstein's picture, radiation travels as: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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