91Ó°ÊÓ

For a lens of focal length \(f\) and with the object distance \(p\) and image distance \(q\), we can use the lens formula given by: $$ \frac{1}{f}=\frac{1}{p}+\frac{1}{q} $$ Since the image is formed at the near point, \(q = N\). We need to find the object distance \(p\). Therefore, the lens formula becomes: $$ \frac{1}{f}=\frac{1}{p}+\frac{1}{N} $$ Solve the equation for \(p\): $$ p = \frac{fN}{N-f} $$

Let's call the image height \(h'\). Using similar triangles, we find the magnification \(M\): $$ M=\frac{h'}{h}=\frac{N}{p} = -\frac{N}{N-f} $$ Since the given formula for \(\theta\) is: $$ \theta=\frac{h(N+f)}{N f} $$ We can rewrite it in terms of the image height \(h'\) using the magnification \(M\): $$ \theta=\frac{h'}{Nf}=\frac{Nh}{Nf}=-\frac{h'}{f} $$

The angular magnification is given by the ratio of the angular size of the image to the angular size of the object when viewed directly: $$ M_{\text{angular}} = \frac{\theta_{\text{image}}}{\theta_{\text{object}}} $$ Since \(\theta_{\text{image}} = -\frac{h'}{f}\) and \(\theta_{\text{object}} = \frac{h}{N}\), substituting these in the formula for angular magnification, we get: $$ M_{\text{angular}} = -\frac{\frac{h'}{f}}{\frac{h}{N}} = \frac{h'N}{fh} $$ Using the given formula for angular magnification when the virtual image is at infinity: $$ M_{\text{infinity}} = 1 + \frac{N}{f} $$ Now we can compare the angular magnification for the near point and infinity. In this particular case, the magnification at near point is higher, given that the formula for \(M_{\text{angular}}\) contains extra terms, thus magnification will be higher than \(M_{\text{infinity}}\).