Question

The radius of curvature of a concave mirror is measured by a spherometer is given by $R=\dfrac{{{l}^{2}}}{6h}+\dfrac{h}{2}$. The measured value of $l$ is $3cm$ using a meter scale with least count $0.1cm$ and measured value of $h$ is $0.045cm$ using a spherometer with least count\[0.005cm\]. Compute the relative error in measurement of radius of curvature.
A) $3$
B) $0.3$
C) $0.2$
D) $0.6$

Answer 1

Hint:The distance from the vertex of the spherical mirror to the center of curvature is known as the radius of curvature (represented by$R$). The radius of curvature is the radius of the sphere from which the mirror was cut.

Complete Step-by-step solution:
The radius of curvature of concave mirror is given as $R=\dfrac{{{l}^{2}}}{6h}+\dfrac{h}{2}$
On differentiating both the sides of the equation, we get
$\Delta R = \Delta {\text{ (}}\dfrac{{{l^2}}}{{6h}} + \dfrac{h}{2})$
 $ =\dfrac{6h\left( \Delta \left( {{l}^{2}} \right) \right)-\left( {{l}^{2}} \right)\left( \Delta \left( 6h \right) \right)}{{{\left( 6h \right)}^{2}}}+\dfrac{1}{2}\left( \Delta h \right) $
 $ =\dfrac{6h\left( 2l\left( \Delta l \right) \right)-6{{l}^{2}}\left( \Delta h \right)}{{{\left( 6h \right)}^{2}}}+\dfrac{\left( \Delta h \right)}{2} $
 $ =\dfrac{12hl\left( \Delta l \right)-6{{l}^{2}}\left( \Delta h \right)}{{{\left( 6h \right)}^{2}}}+\dfrac{\left( \Delta h \right)}{2} $
 $ =\dfrac{2hl\left( \Delta l \right)-{{l}^{2}}\left( \Delta h \right)}{6{{h}^{2}}}+\dfrac{\left( \Delta h \right)}{2} $
The measured value of $l=3cm$
The least count for the measurement of $l$ is given as, $\Delta l=0.1cm$
The measured value of $h=0.045cm$
The least count for the measurement of $h$ is given as, $\Delta h=0.005cm$
The relative error can be calculated as,
$ \text{Relative error}=\dfrac{\Delta R}{R}$
 $=2\left( \dfrac{\Delta l}{l} \right)+\left( \dfrac{\Delta h}{h} \right)+\left( \dfrac{\Delta h}{h} \right)
$
Putting the value in the expression for relative error, we get
$ \text{Relative error}=2\left( \dfrac{0.1}{3} \right)+\left( \dfrac{0.005}{0.045} \right)+\left( \dfrac{0.005}{0.045} \right) $
$ =0.2888
$
Rounding off to the nearest two decimals,
The relative error is $0.3$

Hence, option B is the correct answer.

Note:The relative error is not the same as the absolute error.
The relative error is calculated error with relative to the measured value.
In the calculation of relative error we have to put only the magnitude of the errors.