Question

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

• 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 1

Answer: (B), (D)

the absolute error in $d$ increases
the fractional error in $d$ decreases

Given $2dsin\theta =\lambda$

$\therefore d=\frac{\lambda }{2}cosec\theta$ .....(i)

Differentiating the above equation w.r.t. ${}^{\prime }{\theta }^{\prime }$ we get

$\frac{d\left(d\right)}{d\theta }=\frac{\lambda }{2}[-cosec\theta cot\theta ]$

$\therefore d\left(d\right)=-\frac{\lambda }{2}cosec\theta cot\theta d\theta$ ......(ii)

On dividing (ii) and (i) we get

$\therefore \left|\frac{d\left(d\right)}{d}\right|=cot\theta d\theta$

As $\theta$ increases from $0^o$ to ${90}^o,cot\theta$ decreases and therefore $\left|\frac{d\left(d\right)}{d}\right|$ decreases $\Rightarrow$ Option (D) is correct

From (ii) $|d\left(d\right)|=\frac{\lambda }{2}\frac{cos\theta }{{sin}^2\theta }$

This value of $\frac{cos\theta }{si{n}^2\theta }$ decreases as $\theta$ increases from $0^o$ to ${90}^o$