Multiple Choice & Assertion-Reason

PHYExam Practice

This is a whole-chapter revision set: the one-mark MCQs and Assertion-Reason items sample every corner of Optics, so they reward knowing the structure of each formula (what is proportional to what) far more than heavy calculation. The questions here cluster in wave optics — fringe width $\beta = \dfrac{\lambda D}{d}$ and how it scales when $d$ , $D$ or the medium changes; single-slit central-maximum width; the coherence condition; the intensity relation $I = 4I_0\cos^2(\phi/2)$ and the $I_{max}:I_{min}$ ratio; Rayleigh resolution $\dfrac{1.22\lambda}{D}$ ; and Malus's and Brewster's laws — but the same proportional-reasoning skill carries over to ray-optics one-liners (mirror/lens formulae, magnification, refraction, TIR, prisms).

Examiners use them to test whether you can reason from a relation and judge cause-and-effect, which is exactly what Assertion-Reason questions probe.

Fringe width and its medium dependence (YDSE)

\beta = \dfrac{\lambda D}{d}, \qquad \beta_{medium} = \dfrac{\beta_{air}}{\mu}

\beta

= fringe spacing,

\lambda

= wavelength in air,

D

= slit-to-screen distance,

d

= slit separation,

\mu

= refractive index of the surrounding medium. In a medium the wavelength shrinks to

\lambda/\mu

, so

\beta

shrinks by the same factor.

Single-slit diffraction: central maximum width

\theta = \dfrac{2\lambda}{a}, \qquad w_0 = \dfrac{2\lambda D}{a}

\theta

= full angular width of the central bright band,

a

= slit width,

D

= slit-to-screen distance,

w_0

= linear width of the central maximum. The central maximum is twice as wide as any secondary maximum, and widens as

a

decreases (width

\propto 1/a

Two-beam interference intensity

I = 4I_0\cos^{2}\!\left(\dfrac{\phi}{2}\right), \qquad \phi = \dfrac{2\pi}{\lambda}\,\Delta x

I_0

= intensity of each beam alone,

\phi

= phase difference,

\Delta x

= path difference. Maxima (

I = 4I_0

) at

\phi = 2n\pi

i.e.

\Delta x = n\lambda

; minima (

I = 0

) at

\phi = (2n-1)\pi

i.e.

\Delta x = (2n-1)\dfrac{\lambda}{2}

, with

n = 1, 2, 3, \dots

Max/min intensity from amplitude or intensity ratio

\dfrac{I_{max}}{I_{min}} = \left(\dfrac{a_1 + a_2}{a_1 - a_2}\right)^{2} = \left(\dfrac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}}\right)^{2}

a_1, a_2

= amplitudes,

I_1, I_2

= intensities of the two beams, with

I \propto a^{2}

. Use the amplitude (square-root of intensity) ratio inside the brackets, not the raw intensity ratio.

Malus's law and Brewster's law

I = I_0\cos^{2}\theta, \qquad \mu = \tan\theta_p, \qquad \theta_p + r = 90^{\circ}

I_0

= intensity of plane-polarised light incident on the analyser,

\theta

= angle between polariser and analyser axes (max at

\theta = 0^{\circ}

, zero at

\theta = 90^{\circ}

\theta_p

= polarising (Brewster) angle,

r

= angle of refraction. At

\theta_p

the reflected and refracted rays are

90^{\circ}

apart.

Rayleigh limit of resolution (telescope)

\Delta\theta = \dfrac{1.22\,\lambda}{D}, \qquad R = \dfrac{1}{\Delta\theta} = \dfrac{D}{1.22\,\lambda}

\Delta\theta

= smallest resolvable angular separation,

D

= objective (aperture) diameter,

\lambda

= wavelength. Smaller

\Delta\theta

means better resolution, so a larger aperture and shorter wavelength both improve resolving power

R

Coherence condition: two sources are coherent only if they emit waves of the same frequency (hence the same wavelength in a given medium) with a constant phase difference. Same amplitude or same intensity is neither necessary nor sufficient; two independent bulbs never interfere because their phase difference fluctuates randomly.
Reason out fringe-width changes from $\beta = \dfrac{\lambda D}{d}$ rather than memorising cases: doubling $d$ and halving $D$ together divide $\beta$ by $4$ ; immersing the apparatus in a medium of index $\mu$ multiplies $\beta$ by $1/\mu$ because $\lambda$ becomes $\lambda/\mu$ (the geometry $D$ , $d$ is unchanged).
Single-slit diffraction geometry: the central maximum spans from the first minimum on one side to the first on the other, giving it twice the width of any secondary maximum. Its angular width $\propto 1/a$ , so a narrower slit produces a broader, more spread-out central band.
Distinguish the two-slit ( $d \sin\theta = n\lambda$ for maxima, $n = 0, 1, 2, \dots$ ) from single-slit ( $a \sin\theta = n\lambda$ for minima, $n = 1, 2, 3, \dots$ and never $n = 0$ , which is the central maximum) conditions: in interference $n\lambda$ marks bright fringes, but in single-slit diffraction $n\lambda$ marks the dark minima — opposite roles for the same right-hand side.
Intensity at a point follows $I = 4I_0\cos^{2}(\phi/2)$ : a path difference of $\lambda/2$ gives $\phi = \pi$ , so $\cos^{2}(\pi/2) = 0$ and the intensity is zero (destructive). A path difference of $\lambda$ gives $\phi = 2\pi$ and full intensity $4I_0$ .
For $I_{max}:I_{min}$ always convert an intensity ratio to an amplitude ratio first ( $a \propto \sqrt{I}$ ): an intensity ratio $1:4$ gives amplitudes $1:2$ , so $I_{max}:I_{min} = (1+2)^{2}:(2-1)^{2} = 9:1$ .
Polarisation proves light is a transverse wave. Malus's law $I = I_0\cos^{2}\theta$ peaks when polariser and analyser axes are parallel ( $\theta = 0^{\circ}$ ) and vanishes when crossed ( $\theta = 90^{\circ}$ ); when unpolarised light of intensity $I_0$ first enters a single polaroid, the output is $I_0/2$ .
Assertion-Reason method: judge the Assertion's truth and the Reason's truth independently, then decide whether the Reason is the correct explanation of the Assertion — a true Reason that simply restates a different fact does not 'explain' the Assertion (e.g. central-maximum width $\propto 1/a$ both states the fact and explains why a narrower slit widens it).

Where the marks go

Treating 'same amplitude/intensity' as the test for coherence. The defining requirement is same frequency plus a constant (time-independent) phase difference; equal intensities only improve fringe contrast, they do not create coherence.
Using the raw intensity ratio inside the $I_{max}:I_{min}$ bracket instead of the amplitude (square-root) ratio. For $I_1:I_2 = 1:4$ the correct answer is $9:1$ , not the wrong $25:9$ you get by plugging in $1$ and $4$ directly.
Swapping the maxima/minima conditions between interference and single-slit diffraction: $n\lambda$ gives bright fringes in two-slit interference but dark minima in single-slit diffraction (with $n = 1, 2, 3, \dots$ , never $n = 0$ ). Also forgetting the central maximum is twice as wide as the side maxima.
Malus/Brewster slips: thinking Malus's intensity is maximum at $90^{\circ}$ (it is maximum at $0^{\circ}$ ), and forgetting that at the polarising angle the reflected and refracted rays are perpendicular ( $90^{\circ}$ apart), which is what makes $\mu = \tan\theta_p$ .

How the board asks it

Assertion–Reasonfringe width and the medium
Assertion: When the whole Young's double-slit apparatus is immersed in water, the fringe width decreases. Reason: The wavelength of light decreases in water, while $\beta = \dfrac{\lambda D}{d}$ keeps $D$ and $d$ unchanged. Choose: (a) both true and Reason is the correct explanation; (b) both true but Reason is not the explanation; (c) Assertion true, Reason false; (d) Assertion false, Reason true.
Assertion–Reasonthe coherence condition
Assertion: Two independent light bulbs placed side by side cannot produce a sustained interference pattern. Reason: They emit light waves of the same intensity. Select the correct option from (a)-(d) of the standard Assertion-Reason key.
Give reasons $\beta = \dfrac{\lambda D}{d}$ proportionality
In a Young's double-slit experiment the slit separation $d$ is doubled and the screen distance $D$ is halved, the source being unchanged. Calculate the factor by which the fringe width $\beta$ changes.
Numericalmax/min intensity from amplitude ratio
Two coherent sources have intensities in the ratio $1:4$ . Calculate the ratio $I_{max}:I_{min}$ in the resulting interference pattern.
Applicationmalus's law with a single polaroid
Unpolarised light of intensity $I_0$ passes first through one polaroid and then through a second polaroid whose axis is at $60^\circ$ to the first. Calculate the intensity of the emergent light in terms of $I_0$ .
Distinguishtwo-slit vs single-slit conditions
For the condition $a\sin\theta = n\lambda$ in single-slit diffraction, state whether $n\lambda$ marks a maximum or a minimum, and explain how this role differs from that of $d\sin\theta = n\lambda$ in double-slit interference.

PreviousResolving Power

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.