Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Wave Optics question

2019 · Shift 2 · Q42
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Physics
  4. /Wave Optics
  5. /2019 · Shift 2 · Q42

Wave Optics question

2019 · Shift 2 · Q42

JEE AdvancedPhysicsWave OpticsMultiple correct+4 / −1
In a Young's double slit experiment, the slit separation d is 0.3 mm and the screen distance D is 1 m. A parallel beam of light of wavelength 600 nm is incident on the slits at angle α\alphaα as shown in figure. JEE Advanced 2019 Paper 2 Offline Physics - Wave Optics Question 17 English On the screen, the point O is equidistant from the slits and distance PO is 11.0 mm. Which of the following statement(s) is/are correct?
  1. A
    For α\alphaα = 0, there will be constructive interference at point P.
  2. B
    For α=0.36π\alpha = {{0.36} \over \pi }α=π0.36​ degree, there will be destructive interference at point P.
  3. C
    For α\alphaα=α=0.36π\alpha = {{0.36} \over \pi }α=π0.36​ degree, there will be destructive interference at point O.
  4. D
    Fringe spacing depends on α\alphaα.
View written solutionFree

Correct answer: B

  1. Given data
  • Slit separation: d=0.3 mm=3×10−4 md = 0.3\text{ mm} = 3\times 10^{-4}\text{ m}d=0.3 mm=3×10−4 m
  • Screen distance: D=1 mD = 1\text{ m}D=1 m
  • Wavelength: λ=600 nm=6×10−7 m\lambda = 600\text{ nm} = 6\times 10^{-7}\text{ m}λ=600 nm=6×10−7 m
  • Distance PO=11.0 mm=1.1×10−2 mPO = 11.0\text{ mm} = 1.1\times 10^{-2}\text{ m}PO=11.0 mm=1.1×10−2 m

In YDSE, fringe width is

β=Dλd\beta = \frac{D\lambda}{d}β=dDλ​

So,

β=1⋅6×10−73×10−4=2×10−3 m=2 mm\beta = \frac{1\cdot 6\times 10^{-7}}{3\times 10^{-4}} = 2\times 10^{-3}\text{ m} = 2\text{ mm}β=3×10−41⋅6×10−7​=2×10−3 m=2 mm

Thus point PPP is at

POβ=112=5.5\frac{PO}{\beta} = \frac{11}{2} = 5.5βPO​=211​=5.5

fringe widths from OOO.

  1. Path difference with oblique incidence

If the incident beam makes angle α\alphaα, then there is an initial path difference at the slits:

Δi=dsin⁡α\Delta_i = d\sin\alphaΔi​=dsinα

At a point at transverse distance yyy on the screen, the geometric path difference due to position is

Δg=dyD\Delta_g = \frac{dy}{D}Δg​=Ddy​

Hence net path difference is

Δ=dyD−dsin⁡α\Delta = \frac{dy}{D} - d\sin\alphaΔ=Ddy​−dsinα

(up to sign convention; only condition for maxima/minima matters).

Therefore,

  • Bright fringe: Δ=nλ\Delta = n\lambdaΔ=nλ
  • Dark fringe: Δ=(n+12)λ\Delta = \left(n+\tfrac12\right)\lambdaΔ=(n+21​)λ
  1. Check option A: α=0\alpha=0α=0 at point PPP

For α=0\alpha=0α=0,

Δ=dyD\Delta = \frac{dy}{D}Δ=Ddy​

At PPP, y=11 mmy=11\text{ mm}y=11 mm, so

Δ=(3×10−4)(11×10−3)1=3.3×10−6 m\Delta = \frac{(3\times10^{-4})(11\times10^{-3})}{1} = 3.3\times10^{-6}\text{ m}Δ=1(3×10−4)(11×10−3)​=3.3×10−6 m

Now,

Δλ=3.3×10−66×10−7=5.5\frac{\Delta}{\lambda} = \frac{3.3\times10^{-6}}{6\times10^{-7}} = 5.5λΔ​=6×10−73.3×10−6​=5.5

So,

Δ=5.5λ=(5+12)λ\Delta = 5.5\lambda = \left(5+\tfrac12\right)\lambdaΔ=5.5λ=(5+21​)λ

This is destructive, not constructive.

So A is false.

  1. Evaluate α=0.36π\alpha = \dfrac{0.36}{\pi}α=π0.36​ degree

Convert to radians:

α=0.36π⋅π180=0.36180=0.002 rad\alpha = \frac{0.36}{\pi}\cdot \frac{\pi}{180} = \frac{0.36}{180} = 0.002\text{ rad}α=π0.36​⋅180π​=1800.36​=0.002 rad

Since angle is small,

sin⁡α≈0.002\sin\alpha \approx 0.002sinα≈0.002

Thus,

dsin⁡α=3×10−4×0.002=6×10−7 m=λd\sin\alpha = 3\times10^{-4}\times 0.002 = 6\times10^{-7}\text{ m} = \lambdadsinα=3×10−4×0.002=6×10−7 m=λ

So oblique incidence introduces path difference of exactly one wavelength.

  1. Check option B: destructive interference at PPP

At PPP,

Δg=dyD=5.5λ\Delta_g = \frac{dy}{D} = 5.5\lambdaΔg​=Ddy​=5.5λ

Net path difference:

Δ=5.5λ−λ=4.5λ\Delta = 5.5\lambda - \lambda = 4.5\lambdaΔ=5.5λ−λ=4.5λ

Since

4.5λ=(4+12)λ4.5\lambda = \left(4+\tfrac12\right)\lambda4.5λ=(4+21​)λ

it is destructive interference.

So B is true.

  1. Check option C: destructive interference at OOO

At point OOO, y=0y=0y=0, so

Δ=−dsin⁡α=−λ\Delta = -d\sin\alpha = -\lambdaΔ=−dsinα=−λ

Magnitude of path difference is λ\lambdaλ, i.e. an integral multiple of wavelength. Hence interference is constructive, not destructive.

So C is false.

  1. Check option D: fringe spacing depends on α\alphaα

Position of bright fringes is given by

dyD−dsin⁡α=nλ\frac{dy}{D} - d\sin\alpha = n\lambdaDdy​−dsinα=nλ

So,

yn=Dsin⁡α+nDλdy_n = D\sin\alpha + n\frac{D\lambda}{d}yn​=Dsinα+ndDλ​

This shows that α\alphaα only shifts the whole fringe pattern by Dsin⁡αD\sin\alphaDsinα. The spacing between consecutive fringes is still

β=Dλd\beta = \frac{D\lambda}{d}β=dDλ​

which is independent of α\alphaα.

So D is false.

  1. Final conclusion

Only option B is correct.

PreviousNext

More from Wave Optics

  • Two coherent monochromatic point sources S1​ and S2​ of wavelength λ=600nm are placed symmetrically on either side of the center of the circle as shown. The sources are separated by a distance d=1.8mm. This… Includes diagram2017 · Multiple correct
  • While conducting the Young's double slit experiment, a student replaced the two slits with a large opaque plate in the XY-plane containing two small holes that act as two coherent point sources (S1, S2) emitting light of wavelength 600 mm.… Includes diagram2016 · Multiple correct
  • A Young's double slit interference arrangement with slits S1 and S2 is immersed in water (refractive index = 4/3) as shown in the figure. The positions of maxima on the surface of water are given by x2 = p2m2 λ 2 − d2, where λ… Includes diagram2015 · Numerical
  • A light source, width emits two wavelengths λ 1 = 400 nm and λ 2 = 600 nm, is used in a Young's double-slit experiment. If recorded fringe widths for λ 1 and λ 2 are β 1 and β 2 and the number of…2014 · Multiple correct
  • In the Young's double-slit experiment using a monochromatic light of wavelength λ, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is2013 · MCQ
  • Young's double slit experiment is carried out by using green, red and blue light, one colour at a time. The fringe widths recorded are β G, β R and β B, respectively. Then,2012 · MCQ
  • A light ray travelling in glass medium is incident on glass-air interface at an angle of incidence θ. The reflected (R) and transmitted (T) intensities, both as function of θ, are plotted. The correct sketch is2011 · MCQ
  • Column I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits S 1​ and S 2​. In each of these cases, S 1​ P 0​ = S 2​ P 0​, S 1​ P 1​− S 2​ P 1​=λ/4… Includes table Includes diagram2009 · MCQ