Wave Optics

JEE Main level, single correct, medium

119 of 191

Question 119 of 191

The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is

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Answer: (B) five

For interference maxima, dsin⁡θ=nλd \sin \theta=n \lambda
Here d=2λd=2 \lambda
∴sin⁡θ=n/2\therefore \sin \theta=\mathrm{n} / 2 and is satisfied by 5 integral values of n(−2,−1,0,1,2)\mathrm{n}(-2,-1,0,1,2), as the maximum value of sin⁡θ\sin \theta can only be 1 .

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