\(M = \dfrac{\beta}{\alpha}\). Here \(\beta\) is the angle the final image subtends at the eye. \(\alpha\) is the angle the object subtends when placed at the least distance of distinct vision \(D\). The size of the image on the retina depends on the angle the object subtends, not on its actual size. That angle decides how large a thing looks.
The Workbook
Questions on Optical Instruments and the Eye, the kind your examinations ask. Solve one on paper. Then tell the book how it went. The book believes you, and it uses your answer to plan your revision.
Microscopes, Telescopes and Defects of Vision
24 questions · after Lesson 5, Telescopes
Simple microscope
Image at \(D\): \(M = 1 + \dfrac{D}{f} = 1 + \dfrac{25}{5} = 6\). Image at infinity (relaxed eye): \(M = \dfrac{D}{f} = 5\).
- Find the object distance.
- Why must the object lie inside \(f\)?
\(v = -25\), \(f = 5\): \(\dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f} = -\dfrac{1}{25} - \dfrac{1}{5} = -\dfrac{6}{25}\), so \(u = -\dfrac{25}{6} \approx -4.2\) cm.
Only an object inside the focal length gives a virtual, erect and enlarged image on the same side, which the eye can view directly.
\(M = 1 + \dfrac{D}{f}\) grows as \(f\) shrinks. But a very short \(f\) needs strongly curved faces of small radius, which brings large spherical and chromatic aberration and a tiny lens. In practice a single lens is limited to about \(10\) times.
At \(D\): \(1 + \dfrac{25}{2.5} = 11\). Relaxed eye: \(\dfrac{25}{2.5} = 10\).
Compound microscope
- Find the position and magnification of the intermediate image \(I_1\).
- Find the magnifying power of the eyepiece and of the microscope.
- Find the tube length \(L\).
\(u_o = -1.1\), \(f_o = 1\): \(\dfrac{1}{v_o} = 1 - \dfrac{1}{1.1} = \dfrac{0.1}{1.1}\), so \(v_o = 11\) cm. \(m_o = \dfrac{v_o}{u_o} = -10\).
Eyepiece with image at \(D\): \(M_e = 1 + \dfrac{25}{5} = 6\). Total \(M = m_oM_e = -60\).
For the eyepiece, \(v_e = -25\): \(\dfrac{1}{u_e} = -\dfrac{1}{25} - \dfrac{1}{5}\), so \(u_e = -\dfrac{25}{6} \approx -4.2\) cm. \(L = v_o + |u_e| = 11 + 4.2 \approx 15.2\) cm.
The objective's magnification \(\dfrac{v_o}{u_o}\) is large only when the object is placed just outside a very short focal length, which also keeps the tube short. The eyepiece works as a simple microscope; a moderate focal length gives a comfortable, aberration-free virtual image for the eye.
\(M = m_o \times M_e = (-20) \times 6 = -120\); the minus sign records the inversion.
Final image at infinity, so \(I_1\) is at the focus of the eyepiece: \(v_o = 15 - 4 = 11\) cm. \(\dfrac{1}{u_o} = \dfrac{1}{11} - \dfrac{1}{1} = -\dfrac{10}{11}\), so \(u_o = -1.1\) cm. \(m_o = \dfrac{11}{-1.1} = -10\); \(M_e = \dfrac{25}{4} = 6.25\); \(M = -62.5\).
The object lies just outside the objective's focus, so the objective forms a real, inverted, enlarged image inside the tube. That image falls inside the focal length of the eyepiece. The eyepiece therefore acts as a magnifying glass and gives a virtual final image for the eye to view.
\(M_e = \dfrac{D}{f_e} = 5\), so \(M = -10 \times 5 = -50\). \(I_1\) now sits at the eyepiece focus: \(L = v_o + f_e = 11 + 5 = 16\) cm.
Telescopes
Normal adjustment: \(M = \dfrac{f_o}{f_e} = \dfrac{100}{5} = 20\) and \(L = f_o + f_e = 105\) cm.
- Why do the foci of the two lenses coincide?
- Why is the objective made with a large aperture?
Parallel rays from a distant star form \(I_1\) at the objective's focus. For the eyepiece to send out parallel rays, which the relaxed eye focuses, \(I_1\) must lie at the eyepiece's focus too. So \(F_o\) and \(F_e\) coincide.
A large objective gathers more light, so faint objects become visible, and its larger aperture gives finer resolution of detail.
\(M = \dfrac{f_o}{f_e}\left(1 + \dfrac{f_e}{D}\right) = 20 \times 1.2 = 24\). Eyepiece: \(v_e = -25\), so \(\dfrac{1}{u_e} = -\dfrac{1}{25} - \dfrac{1}{5}\) and \(u_e = -\dfrac{25}{6} \approx -4.2\) cm. \(L = 100 + 4.2 \approx 104.2\) cm.
The erecting lens is placed so that \(I_1\) is at \(2f\) on one side and its erect image at \(2f\) on the other. This adds \(4f = 20\) cm: \(L = 105 + 20 = 125\) cm. Without it the astronomical telescope shows everything upside down, which is harmless for stars but useless on land.
A mirror focuses all colours alike, so there is no chromatic aberration. A large mirror can be supported over its whole back. A large lens sags under its own weight and must be held by its rim. A parabolic mirror is free of spherical aberration, and light does not pass through glass, so absorption is avoided.
The telescope multiplies the angle by \(M = 20\): \(0.5^\circ \times 20 = 10^\circ\). An astronomical telescope gives an inverted image.
Light from the star enters and strikes the large concave primary mirror, which reflects it towards its focus. Before reaching the focus it meets the small convex secondary mirror. That mirror sends it back through a hole in the centre of the primary to the eyepiece behind. The folded path makes the instrument short.
Defects of vision
- Where do parallel rays focus, and why can the eye not see distant objects clearly?
- Find the focal length and power of the correcting lens.
They focus in front of the retina: the eye's lens is too strong, or the eyeball too long. Only objects nearer than \(2\) m form sharp images on the retina.
The lens must make distant objects appear at the far point: a virtual image at \(2\) m for an object at infinity. So \(f = -2\) m, a concave lens of power \(P = \dfrac{1}{-2} = -0.5\) D.
Hypermetropia (long sight). The lens must form a virtual image at the near point, \(50\) cm, of an object at \(25\) cm. \(\dfrac{1}{f} = \dfrac{1}{-50} - \dfrac{1}{-25} = \dfrac{1}{50}\), so \(f = +50\) cm, a convex lens of \(+2\) D.
Presbyopia is the loss of the eye's power of accommodation with age: the near point recedes, and often distant vision weakens too. The eye lens stiffens and the ciliary muscles weaken, so the lens can no longer bulge enough. It is corrected with bifocal spectacles: a convex part for reading and, if needed, another part for distance.
In astigmatism the cornea (or lens) is curved differently in different planes. Lines in one direction focus sharply while lines at right angles are blurred. A cylindrical lens, which has power in one plane only, is used to make the total curvature the same in all planes.
Distant vision: \(f = -50\) cm, \(P = \dfrac{1}{-0.5} = -2\) D. With this lens, the nearest object is one whose virtual image falls at the near point \(12\) cm: \(\dfrac{1}{u} = \dfrac{1}{-12} - \dfrac{1}{-50} = -\dfrac{38}{600}\), so \(u \approx -15.8\) cm. Reading is slightly harder with the lens on.
The lens must image an object at \(25\) cm as a virtual image at the near point \(N\): \(u = -25\), \(v = -N\). \(\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} = -\dfrac{1}{N} + \dfrac{1}{25} = \dfrac{N - 25}{25N}\), so \(f = \dfrac{25N}{N - 25}\), positive for \(N > 25\): a convex lens.