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` remain constant.
D. the fractional error in `d` decreases.

Answer 1

Correct Answer - D
`d = ( lambda)/( 2 sin theta) In d = In ((lambda)/( 2)) - In sin theta ( Delta d)/( d) = 0 - ( cos theta d theta)/( sin theta)`
`(( Delta d)/(d))_(max) = +- cot theta Delta theta`
` ( lambda)/( 2 sin theta) cot theta Delta theta = ( lambda)/(2) ( cos theta)/( sin^(2) theta) Delta theta`
As `theta` increases , `cot theta` decreases and `(cos theta)/( sin^(2) theta)` also decreases.