$A$ single slit diffraction experiment is performed to determine the slit width using the equation,$\frac{b d}{D} = m \lambda$,where $b$ is the slit width,$D$ is the distance between the slit and the screen,$d$ is the distance between the $m^{\text{th}}$ diffraction maximum and the central maximum,and $\lambda$ is the wavelength. $D$ and $d$ are measured with scales of least count of $1 \ cm$ and $1 \ mm$,respectively. The values of $\lambda$ and $m$ are known precisely to be $600 \ nm$ and $3$,respectively. The absolute error (in $\mu m$) in the value of $b$ estimated using the diffraction maximum that occurs for $m=3$ with $d=5 \ mm$ and $D=1 \ m$ is $.....$

• A
  $(45.60 \text{ or } 50.50)$
• B
  $(71.60 \text{ or } 60.50)$
• C
  $(76.60 \text{ or } 91.50)$
• D
  $(75.60 \text{ or } 94.50)$