Using the expression `2d sin theta = lambda` , one calculates the values of `d` by measurinf the corresponding angles `theta` in the range ` 0^(@) to 90^(@)`. The wavelength `lambda` is exactly known and the error in `theta` is constant for all values of ` theta` is constant for all the values of `theta` . As `theta` increases from `0^(@)` ,

Options

A. the absolute error in `d` remains constant.
B. the absolute error in `d` increses.
C. the fractional error in `d` remains constant.
D. the fractional error in `d` decreases.

Answer

D

Explanation

`2d sintheta=lambda`
`d=(lambda)/(2 sin theta)`
differntiate
`del(d)=(lambda)/(2) del (cosectheta)`
`del(d)=(lambda)/(2)(-cos esthetacottheta)deltheta`
`del(d)=(-lambdacostheta)/(2sin^(2)theta)deltheta`
as `theta= "increases", (lambdacostheta)/(2sin^(2)theta) , "decreases"`
Alternate solution
`d=(lambda)/(2sintheta)`
`ln d=ln lambda-ln2-ln sintheta`
`(Delta(d))/(d)=0-0-(1)/(sin theta)xxcostheta(Deltatheta)`
Fraction error `|+(d)|=|cotthetaDeltatheta|`
Absoulute error `Deltad=(dcottheta)Deltatheta`
`(d)/(2sintheta)xx(costheta)/(sintheta)`
`Deltad=(costheta)/(sin^(2)theta)`