Optics 6 Question 35

37. Using the expression 2dsinθ=λ, one calculates the values of d by measuring the corresponding angles θ in the range 0 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

(a) the absolute error in d remains constant(2013 Adv.)

(b) the absolute error in d increases

(c) the fractional error in d remains constant

(d) the fractional error in d decreases

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Solution:

  1. d=λ2sinθlnd=lnλln2lnsinθ

Δ(d)d=001sinθ×cosθ(Δθ)

Fractional error =|Δdd|=(cotθ)Δθ

Absolute error Δd=(dcotθ)Δθ

=λ2sinθcosθsinθΔθ

Now, given that Δθ= constant As θ increases, sinθ increases, cosθ and cotθ decrease.

Both fractional and absolute errors decrease.



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