NEET Physics Wave Optics Questions Solved

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(a) Derive an expression for path difference in Young's double slit experiment and obtain  the conditions for constructive and destructive at a point on the screen.

(b) The intensity at the central maxima in Young's double slit experimeny is I0. Find out the intensity at a point where the path difference is λ6,λ4 and λ3

(a)                                                            (1/2) marks

Path difference =S2P-S1P

Now (S2P)2-(S1P)2=D2+x+d22-D2+x+d22


whereS1S2=d and OP=x


For x<<D and d<<D, we can write                                                   (1/2) marks

S2P+S1P2D                                                                               (1/2) marks

Hence, Path difference = S2P-S1P=2xd2D=xdD                                (1/2) marks

For constructive interference, we must have

xdD=                                                                                         (1/2) marks

x=xn=nλDd (n=0,±1,±2.....)                                                     (1/2) marks

For destructive interference , we must have

xdD=n+12λ                                                                                (1/2) marks

x=xn'=n+12λDd (n=0,±1,±2.....)

(b) The general expression, for the intensity, at a point is I=I0cos2ϕ2

       (i) For path difference = λ6            ϕ=60°                               (1/2) marks


       (ii) For path difference = λ4           ϕ=90°                                (1/2) marks


       (iii) For path difference = λ6           ϕ=120°                              (1/2) marks


Difficulty Level:

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