Using the expression 2 d sin theta = lambda,o | vedclass

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Question

(D) Given the expression: $2 d \sin 	heta = \lambda$,we have $d = \frac{\lambda}{2 \sin 	heta}$.Differentiating both sides with respect to $	heta$

Vedclass · Accepted Answer

the fractional error in $d$ decreases.

Vedclass · Answer

(D) Given the expression: $2 d \sin 	heta = \lambda$,we have $d = \frac{\lambda}{2 \sin 	heta}$.Differentiating both sides with respect to $	heta$ to find the absolute error $\Delta d$:$\Delta d = \left| \frac{d}{d	heta} \left( \frac{\lambda}{2 \sin 	heta} ight) ight| \Delta 	heta = \left| -\frac{\lambda \cos 	heta}{2 \sin^2 	heta} ight| \Delta 	heta = \frac{\lambda \cos 	heta}{2 \sin^2 	heta} \Delta 	heta$.Since $\Delta 	heta$ is constant,as $	heta$ increases from $0^{\circ}$ to $90^{\circ}$,$\cos 	heta$ decreases and $\sin 	heta$ increases,so $\Delta d$ decreases.Now,calculating the fractional error $\frac{\Delta d}{d}$:$\frac{\Delta d}{d} = \frac{\frac{\lambda \cos 	heta}{2 \sin^2 	heta} \Delta 	heta}{\frac{\lambda}{2 \sin 	heta}} = \frac{\cos 	heta}{\sin 	heta} \Delta 	heta = \cot 	heta \Delta 	heta$.As $	heta$ increases from $0^{\circ}$ to $90^{\circ}$,$\cot 	heta$ decreases. Therefore,the fractional error $\frac{\Delta d}{d}$ decreases.

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