Discussion Forum

Using the expression   $2d \sin\theta=\lambda$, one calculates the values of d by measuring the corresponding angle $\theta$ in the range 0 to 90°.  The wavelength $\lambda$  is exactly known and the error in $\theta$  is constant for all values of $\theta$. As $\theta$  increases from 0°  

Answer:

Explanation:

$2d\sin\theta=\lambda$

 $d=\frac{\lambda}{2\sin\theta}$

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

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

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

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