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ISC Class 12 Physics — practice by topic

Browse 624 questions across 10 Physics topics. Choose what you want to revise, then open a question to check its answer and worked solution.

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Physics topics

Moving Charges and Magnetism (133)Current Electricity (67)Electromagnetic Induction & Alternating Current (58)Semiconductor Electronics (58)Ray Optics and Optical Instruments (55)Electromagnetic Waves (52)Atoms and Nuclei (51)Dual Nature of Radiation and Matter (50)Electric Charges and Fields (50)Wave Optics (50)

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Unpolarised light of intensity $I_0$ is passed through a single ideal polariser. The intensity of the transmitted light is:Read the answer and worked solution →Brewster's angle for a glass of refractive index $\sqrt{3}$ (relative to air) is:Read the answer and worked solution →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 →

Worked practice questions

Unpolarised light of intensity $I_0$ is passed through a single ideal polariser. The intensity of the transmitted light is:

  1. $I_0$
  2. $I_0/2$
  3. $I_0\cos^2\theta$
  4. $I_0/4$
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Correct answer: B. $I_0/2$

Unpolarised light contains all polarisation directions with equal probability. Averaging Malus's law over all input angles gives $\langle I \rangle = I_0 \langle\cos^2\theta\rangle = I_0/2$. Once polarised, a second polariser would then attenuate further by $\cos^2\theta$ relative to its axis.

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Brewster's angle for a glass of refractive index $\sqrt{3}$ (relative to air) is:

  1. $30^\circ$
  2. $45^\circ$
  3. $60^\circ$
  4. $90^\circ$
Show answer and solution

Correct answer: C. $60^\circ$

**Brewster's law:** $\tan i_B = \mu$. With $\mu = \sqrt{3}$, $\tan i_B = \sqrt{3} \Rightarrow i_B = 60^\circ$. At this angle the reflected light is completely plane-polarised, with the electric field perpendicular to the plane of incidence.

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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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