Question

Using the relation \[m = \frac{v}{u} = v\left(\frac{1}{v} - \frac{1}{f}\right)\] and the sign convention where \[v\] is negative and equal in magnitude to \[D\], what is the linear magnification \[m\] for the image formed at the near point by a simple microscope?

पर्याय

• \[m = \left(1 - \frac{D}{f}\right)\]

• \[m = \left(1 + \frac{D}{f}\right)\]

• \[m = \left(\frac{D}{f}\right)\]

• \[m = \left(\frac{f}{D}\right)\]

Answer 1

Starting from \[m = \frac{v}{u} = v\left(\frac{1}{v} - \frac{1}{f}\right) = \left(1 - \frac{v}{f}\right)\], applying the sign convention where \[v\] is negative and equal in magnitude to \[D\] gives \[m = \left(1 + \frac{D}{f}\right)\]. This shows the magnification at the near point is one more than the case where the image is at infinity.