Physics - Wave Optics Question with Solution | TestHub

PhysicsWave OpticsYDSEEasy2 minQB
PhysicsEasynumerical

In YDSE arrangement as shown in the figure, fringes are seen on screen using monochromatic source S having wavelength3000 A∘(in the air).S1 and S2are two slits separated byd=1 mm and D=1 m. Left of slitsS1 and S2medium of refractive indexn1=2is present and to the right ofS1 and S2medium ofn2=32, is present. A thin slab of thickness 't' is placed in front ofS1. The refractive index ofn3of the slab varies with distance from it's starting face as shown in figure.




In order to get central maxima at the centre of the screen, the thickness of the slab (inμm) required is:

Answer:
2.00
Solution:

Path difference,
∆x=n1 SS2+n2S2P-n1SS1+n2S1P-∫0tn3-n2dx 

=n1 SS2-SS1+n2 S2P-S1P-∫0tn3dx-n2t

In order to get central maxima at the centre of the screen



o=2×1×10-322×1+0-2t+3t2

0.5t=1 μm

t=2 μm

Stream:JEESubject:PhysicsTopic:Wave OpticsSubtopic:YDSE
⏱ 2mℹ️ Source: QB

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