Question 1Level 2 · ExamSingle correct
A plane monochromatic wave travelling in glass (refractive index 1.5) meets a plane glass–air interface at an angle of incidence of 30∘. Take c=3.0×108 m/s in air. Consider a strip of the incident plane wavefront whose width, measured along the wavefront in the plane of incidence, is 2.0 mm. Using Huygens' construction, the width of the corresponding refracted wavefront in air, and the time interval between the instants at which the two edges of this strip reach the interface, are ______. Exactly one option is correct.
A2.62 mm and 5.77 ps
B1.53 mm and 3.85 ps
C1.53 mm and 5.77 ps
D2.00 mm and 5.77 ps
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Answer: C
Let the strip meet the interface along a segment of length L. Since the wavefront makes angle i with the interface, L=w/cosi=2.0/cos30∘=2.309 mm.\nThe far edge still has to travel Lsini=wtani=1.155 mm in glass, at speed v1=c/1.5=2.0×108 m/s: Δt=1.155×10−3/(2.0×108)=5.77 ps.\nIn this time the wavelet from the first edge spreads in air to radius cΔt=Lsinr, with sinr=1.5sin30∘=0.75 (Snell's law follows from the construction). The refracted wavefront is the tangent from the far edge to this wavelet; it makes angle r with the interface, so its width is Lcosr=2.309×0.6614=1.53 mm.\nTraps: using wcosi/cosr gives 2.62 mm; using the speed in air for the incident wave gives 3.85 ps.\nDifficulty: HardQuestion 2Level 3 · AdvancedSingle correct
In a Young's double-slit experiment the slits S1 (at y=+d/2) and S2 (at y=−d/2) are separated by d=0.50 mm, the screen is at D=1.0 m, and light of wavelength 600 nm is used. The point O on the screen is equidistant from the slits. Both slits are illuminated coherently and equally, the slits are narrow, and small-angle approximations hold unless stated otherwise. Two transparent films of the same thickness t are then placed, one over each slit: refractive index 1.4 over S1 and 1.7 over S2 (no absorption). The zero-order (zero optical path difference) bright fringe now lies where the 5th-order bright fringe was before. The displacement of the zero-order fringe and the value of t are ______. Exactly one option is correct.
A6.0 mm towards −y; t=10 μm
B6.0 mm towards +y; t=10 μm
C6.0 mm towards −y; t=4.3 μm
D6.0 mm towards +y; t=2.7 μm
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Answer: A
Fringe width β=λD/d=600×10−9×1.0/(0.50×10−3)=1.2 mm, so the 5th-order fringe was at 5β=6.0 mm from O.\nEach film replaces a length t of air, adding optical path (n−1)t. The path via S2 gains 0.7t, via S1 only 0.4t, so the path via S2 is longer by (1.7−1.4)t=0.3t (the 1's cancel).\nAt a point y the geometric path difference is S2P−S1P=yd/D. Zero optical path difference needs S1P−S2P=0.3t, i.e. y=−0.3tD/d<0: the pattern moves towards S2 (−y), the slit with the larger optical path.\n0.3t=5λ=3.0 μm gives t=10 μm.\nTraps: 5λ/(n2−n1+0.4) or 5λ/(n1+n2−2) give the wrong thicknesses; moving the pattern towards the smaller optical path gives the wrong side.\nDifficulty: Very hardQuestion 3Level 3 · AdvancedSingle correct
Two narrow slits lie in the plane x=0 at y=+d/2 (upper slit) and y=−d/2 (lower slit), with d=0.25 mm; a screen is at x=D=1.0 m and O is the point (D,0). A plane monochromatic wave of wavelength 500 nm travels in the direction (cosθ,sinθ,0), i.e. tilted towards +y, with sinθ=2.0×10−3. A thin transparent film of refractive index 1.5 is to be placed over one slit so that the zero-order (zero optical path difference) maximum falls exactly at O. The film must be placed over ______. Exactly one option is correct.
Athe upper slit, with thickness 1.0 μm
Bthe lower slit, with thickness 1.0 μm
Cthe lower slit, with thickness 0.50 μm
Dthe upper slit, with thickness 2.0 μm
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Answer: B
The incident wavefronts are perpendicular to (cosθ,sinθ). A wavefront that passes the lower slit must still travel an extra dsinθ=0.25×10−3×2.0×10−3=0.50 μm to reach the upper slit, so light at the upper slit lags by an optical path dsinθ.\nWithout a film, zero total path difference needs the screen path from the upper slit to be shorter by dsinθ, i.e. yd/D=dsinθ, y=Dsinθ=+2.0 mm: the central maximum lies along the incident direction, on the +y side.\nTo bring it to O (equal screen paths), the lower slit must be delayed by the same 0.50 μm: (n−1)t=0.50 μm, so t=0.50/0.5=1.0 μm over the lower slit.\nTrap: a film on the upper slit adds to the lag and pushes the zero order further up; forgetting the factor (n−1) gives 0.50 μm.\nDifficulty: Very hard