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Question:

Using the expression 2dsinθ=λ, one calculates the values of d by measuring the corresponding angles θ in the range θ to 90°. The wavelength λ is exactly known and the error in θ is constant for all values of θ. As θ increases from 0°, what happens to the error in d?

the absolute error in d remains constant

the fractional error in d remains constant

the fractional error in d decreases

the absolute error in d increases

Solution:

We have 2dsinθ=λ
Let 2d=y
Thus we get, ysinθ=λ ⇒lny+lnsinθ=lnλ
Differentiating both the sides we get
⇒dy/y=−cotθ
⇒|dy/y|=cotθ
As y=2d, we see that the fractional error in d decreases as θ increases.