A simple magnifier gives the maximum angular magnification when it forms a virtual image at the near point of the eye instead of at infinity. For simplicity, assume that the magnifier is right up against the eye, so that distances from the magnifier are approximately the same as distances from the eye. (a) For a magnifier with focal length \(f,\) find the object distance \(p\) such that the image is formed at the near point, a distance \(N\) from the lens. (b) Show that the angular size of this image as seen by the eye is $$ \theta=\frac{h(N+f)}{N f} $$ where \(h\) is the height of the object. [Hint: Refer to Fig. 24.15 .1 (c) Now find the angular magnification and compare it to the angular magnification when the virtual image is at infinity.

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 \(M_{\text{infinity}}\).