Using the expression 2 d = , one calculates the values of d by me… - Vidyadip

MCQ

Using the expression $2 d \sin \theta=\lambda$, one calculates the values of $d$ by measuring the corresponding angles $\theta$ in the range $0$ to $90^{\circ}$. The wavelength $\lambda$ is exactly knowns and the error in $\theta$ is constant for all values of $\theta$. As $\theta$ increases from $0^{\circ}$ :

A
the absolute error in $d$ remains constant.
B
the absolute error in $d$ increases.
C
the fractional error in $d$ remains constant.
✓
the fractional error in $d$ decreases.

✓

Answer

Correct option: D.

the fractional error in $d$ decreases.

d
$2 d \sin \theta=\lambda$

$d =\frac{\lambda}{2 \sin \theta}$

differntiate

$\partial( d )=\frac{\lambda}{2} \partial(\operatorname{cosec} \theta)$

$\partial( d )=\frac{\lambda}{2}(-\operatorname{cosec} \theta \cot \theta) \partial \theta$

$\partial(d)=\frac{-\lambda \cos \theta}{2 \sin ^2 \theta} \partial \theta $

$\text { as } \theta=\text { increases, } \frac{\lambda \cos \theta}{2 \sin ^2 \theta}$ $\text { decreases }$

Alternate solution

$d=\frac{\lambda}{2 \sin \theta} $

$\ell n d=\ell n \lambda-\ell \ln 2-\ell \sin \theta $

$\frac{\Delta(d)}{d}=0-0-\frac{1}{\sin \theta} \times \cos \theta(\Delta \theta) $

$\text { Fractional error }|+(d)|=|\cot \theta \Delta \theta| $

$\text { Absoulute error } \Delta d=(d \cot \theta) \Delta \theta $

$\frac{d}{2 \sin \theta} \times \frac{\cos \theta}{\sin \theta} $

$\Delta d=\frac{\cos \theta}{\sin ^2 \theta}$

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