JEE Advanced 2013 Paper 2 Physics Q1

Physics · Foundations & Measurement

Problem

Using the expression 2dsin⁡θ=λ2d\sin\theta=\lambda, one calculates the values of dd by measuring the corresponding angles θ\theta in the range 0 to 90∘90^\circ. The wavelength λ\lambda is exactly known and the error in θ\theta is constant for all values of θ\theta. As θ\theta increases from 0∘0^\circ,

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