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. (a) The absolute error in `d` remains constant
B. (b) The absolute error in `d` increases
C. ( c ) The fractional error in `d` remains constant
D. ( d ) The fractional error in `d` decreases

Answer

D

Explanation

(d) given `2d sin theta = lambda` :. `d = (lambda)/(2)cosec theta` …(i)
:. `(d(d))/(d theta) = (lambda)/(2)[-cosec theta cot theta]`
:. `d(d) = -(lanbda)/(2) cosec theta cot theta` …(ii)
on dividing (i) and (ii) we get
:. `|(d(d))/(d)| = cot theta d theta`
As `theta` increase from `0@ to 90@, cot theta` decreases and therefore `|(d(d))/(d)|` decrease option `(d)` is correct
From (ii) `|d(d)| = (lambda)/(2)(cos theta)/(sin^(2) theta)`
This value of `(cos theta)/(sin^(2) theta)` decreases from `0@ to 90@`