A microscope has an objective of focal length $2\text{cm}$, eyepiece of focal length $4\text{cm}$ and the tube length of $40\text{cm}$. If the distance of distinct vision of eye is $25\text{cm}$, the magnification in the microscope is:

The total magnification ($m$) of a compound microscope when the final image is formed at the least distance of distinct vision ($D$) is given by:
$$m=m_o\times m_e=\left(\frac{L}{f_o}\right)\left(1+\frac{D}{f_e}\right)$$
Given:Tube length $L=40\text{cm}$ Focal length of objective $f_o=2\text{cm}$ Focal length of eyepiece $f_e=4\text{cm}$ Least distance of distinct vision $D=25\text{cm}$ Substituting the values:
$$m=\left(\frac{40}{2}\right)\left(1+\frac{25}{4}\right)$$ $$m=20\times (1+6.25)=20\times 7.25=145$$
Rounding to the nearest standard option provided in the context of such problems (often using $m\approx \frac{L\cdot D}{f_o\cdot f_e}$ for relaxed eye or specific approximations):Using $m\approx \frac{L}{f_o}\cdot \frac{D}{f_e}=\frac{40}{2}\cdot \frac{25}{4}=20\cdot 6.25=125$.Based on the image options, 125 is the intended answer.