Optics 6 Question 35
37. 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 known 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(2013 Adv.)
(b) the absolute error in $d$ increases
(c) the fractional error in $d$ remains constant
(d) the fractional error in $d$ decreases
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Solution:
- $d=\frac{\lambda}{2 \sin \theta} \ln d=\ln \lambda-\ln 2-\ln \sin \theta$
$$ \frac{\Delta(d)}{d}=0-0-\frac{1}{\sin \theta} \times \cos \theta(\Delta \theta) $$
Fractional error $=\left|\frac{\Delta d}{d}\right|=(\cot \theta) \Delta \theta$
Absolute error $\Delta d=(d \cot \theta) \Delta \theta$
$$ =\frac{\lambda}{2 \sin \theta} \frac{\cos \theta}{\sin \theta} \Delta \theta $$
Now, given that $\Delta \theta=$ constant As $\theta$ increases, $\sin \theta$ increases, $\cos \theta$ and $\cot \theta$ decrease.
$\therefore$ Both fractional and absolute errors decrease.
