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@`

Options

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

D

Explanation

`d = (lamda)/(2 sin theta)`
`delta(d)=((lamda)/(2sin^(2)theta))cos theta delta theta (delta theta ="constant")`
as `theta` increasesm `(cos theta)/(sin^(2)theta)` decrease so
Absolute error `|delta(d)|` decreases
Also fractional error
`=|(delta(d))/(d)|=(((lamdacos theta)/(2 sin^(2)theta))deltatheta)/((lamda)/(2 sin theta))=(cos theta)delta theta`
as `theta` increases, `cot theta` decreases, so fractional error decreases.