Question

Using the expression `2d sin theta = lambda`, one calculates the values of `d` by measuring the corresponding angles `theta` in the range `0 to 90@`. The wavelength `lambda` is exactly known and error in `theta` is constant for all values of `theta`. As `theta` increases from `0@`
A. the absolute error in d remains constant
B. the absolute error in d increases
C. the fractional error in d remains constant
D. the fractional error in d decreases

Answer 1

Correct Answer - d
`d=lambda/(2 sin theta)lnd=ln (lambda/2)-ln sin theta (Deltad)/d=0-(cos theta d theta)/(sin theta)`
`((Deltad)/d)_(max)= pm cos theta Delta theta`
Also `(Deltad)_(max)=d cot theta Delta theta`
`lambda/(2 sin theta) cot theta Delta theta=lambda/2 (cos theta)/(sin^(2) theta)Deltatheta`
As `theta` increases, `cot theta` decreases and `(cos theta)/(sin^(2) theta)` also decreases.