In a special arrangement of Young's double-slit experiment, the distance between the slits d is twice the distance between the screen and the slits D…

Question

In a special arrangement of Young's double-slit experiment, the distance between the slits $d$ is twice the distance between the screen and the slits $D$ , i.e $d = 2 D$ . For this setup, the value of $D$ such that the first minima on the screen fall at a distance $D$ from the centre $O$ is found to be $\frac{λ}{N}$ .

What is the value of $N$ ? [Take $\sqrt{5} = 2.24$ ]

MARKS Content Team · Accepted Answer

Correct answer: 2.48. From diagram as provided in question, OP=x CO=D S_1C=S_2C=DT_1P=T_1O-OP=D-x T_2P=T_2O+OP=D+x Now, S_1p=√(S1T1^2+T1 P^2)=√(D^2+D-x^2) S_2P=√(S2T2^2+T2P^2)=√(D^2+D+x^2) For fist minimum to occur, Path difference S_2P-S_1P=(λ)/2 ⇒√(D^2+D+x^2)-√(D^2+D-x^2)=(λ)/2 The first minimum falls at a distance D from the centre, i.e., x=D D2+4D2^12-D=(λ)/2 ∴D√5-1=(λ)/2 ⇒D=(λ)/(22.24-1)=(λ)/2.48

Solution

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