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

Using the given espression we can write the following :
`d= (lambda)/(2) "cosec" theta`
By differentiating the above relation we can write the following :
`(triangled)/(triangle theta)= -(lambda)/(2) "cosec" theta "cot" theta`
Hence absolute error in d can be written as follows :
`|triangled|=(lambda)/(2) "cosec" theta "cot" theta triangle theta " "".........."(i)`
Fractional error in d can be written as follows :
`|(triangle d)/(d)|=((lambda)/(2) "cosec theta cot theta triangle theta)/((lambda)/(2) "cosec" theta)`
`implies |(triangle d)/(d)|= cot theta triangle theta" "".........."(ii)`
From equation (i) we can see that absolute error in d is not constant when `theta` is varied from `0 "to" 90^(@)`. Hence option a is wrong.
When `theta` is increased from `0 "to"90^(@)` then value of `"cosec" theta cot theta` decreases from infinity to zero hence option b is wrong.
From equation (ii) we can see that fractional error in d changes when `theta` is changed hence option is correct.
When `theta` is increased from `0 "to"90^(@)` then value of `cot theta` decreases hence option d is correct.