JEE Physics Ray Optics and Optical Instruments Notes Set 06

The Human Eye

Light enters through the cornea and the pupil (the hole in the iris, whose size is adjusted by muscles), and the eye lens focuses it as a real, inverted, diminished image on the retina. The retina has rods (sensitive to intensity) and cones (sensitive to colour) that send signals to the brain along the optic nerve. The distance from lens to retina is fixed, about 2.5 cm, so the eye focuses objects at different distances by changing the focal length of the lens with the ciliary muscles. This ability is called accommodation.

1. Near point: the closest distance at which the eye can focus, taken as \(D = 25\) cm for normal vision. Far point: the farthest distance, which is infinity for a normal eye.
2. The eye is most relaxed when it looks at infinity (normal adjustment) and under maximum strain for an object at the near point.
3. The eye is most sensitive to yellow-green light of wavelength about 5550 \(\text{\AA}\).
4. The apparent size of an object depends on the visual angle it subtends at the eye, not just on its actual size.
5. The limit of resolution of the eye is about one minute of arc; two objects subtending less than this cannot be seen as separate.
6. Persistence of vision is about \(\frac{1}{10}\) s, which is why a rapid series of pictures looks like continuous motion.

Defects of Vision and Their Correction

Derivation for myopia: the lens must make an object at infinity appear at the far point, so \(u = -\infty\) and \(v = -x\), giving \(f = -x\). For hypermetropia the lens must make an object at \(-D\) appear at \(-x\), so \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = -\frac{1}{x} + \frac{1}{D}\).

Worked examples: a myopic person who cannot see beyond 80 cm needs \(f = -80\) cm, power \(-1.25\) D. A person whose near point is 50 cm needs \(f = \frac{50 \times 25}{50 - 25} = 50\) cm, power \(+2\) D. With the far point at 150 cm, the lens power is \(-\frac{1}{1.5} = -0.67\) D, and such a person reading through the glasses at 25 cm has the image at \(v = \frac{uf}{u + f} = \frac{(-25)(-150)}{-25 - 150} = -21.4\) cm, so the near point with glasses is 21.4 cm.

Simple Microscope (Magnifier)

A convex lens used with the object inside its focal length forms a virtual, magnified, erect image. Its angular magnification is the ratio of the angle subtended by the image to the angle subtended by the object when placed at the near point.

1. Image at infinity (relaxed eye, object at the focus): \(M = \frac{D}{f}\).
2. Image at the near point: \(M = 1 + \frac{D}{f}\), which is the maximum magnifying power.
3. Object distance range for clear vision with image between \(D\) and infinity: from \(u = -\frac{fD}{D + f}\) to \(u = -f\).

Worked example: a lens of \(f = 5\) cm used by a person with near point 25 cm. The closest object distance is \(\frac{5 \times 25}{30} = 4.17\) cm and the farthest is 5 cm. The magnifying power ranges from \(\frac{25}{5} = 5\) to \(1 + \frac{25}{5} = 6\).

Compound Microscope

It has two convex lenses on a common axis: an objective of very small focal length and aperture facing the object, and an eyepiece of small focal length but larger aperture. The object is placed just outside the focus of the objective, which forms a real, inverted and enlarged image just inside the focus of the eyepiece. The eyepiece acts as a magnifier and gives a final virtual, inverted image.

1. \(M = m_o \times m_e = \frac{v_o}{u_o}\left(1 + \frac{D}{f_e}\right)\) with the final image at the near point.
2. With the final image at infinity: \(M = \frac{v_o}{u_o}\cdot\frac{D}{f_e}\).
3. Approximately, with \(u_o \approx f_o\) and \(v_o \approx L\) (tube length): \(M \approx -\frac{L}{f_o}\left(1 + \frac{D}{f_e}\right)\) (image at near point) or \(-\frac{LD}{f_o f_e}\) (image at infinity). The negative sign means an inverted image.
4. Both focal lengths should be small for high magnification.

Worked example: \(f_o = 2\) cm, \(f_e = 5\) cm, lenses 20 cm apart, final image at 25 cm from the eyepiece. For the eyepiece, \(\frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e} = -\frac{1}{25} - \frac{1}{5}\), so \(u_e = -\frac{25}{6}\) cm. Then \(v_o = 20 - \frac{25}{6} = \frac{95}{6}\) cm, and from the lens formula for the objective, \(u_o = -\frac{190}{83} \approx -2.3\) cm. The magnifying power is \(-\frac{v_o}{|u_o|}\left(1 + \frac{25}{5}\right) \approx -41.5\).

Astronomical Telescope

It has an objective of large focal length and aperture and an eyepiece of small focal length. The objective forms a real inverted image of a distant object at its focal plane, and the eyepiece magnifies it.

1. Normal adjustment (final image at infinity): \(M = -\frac{f_o}{f_e}\) and the tube length is \(L = f_o + f_e\).
2. Final image at the near point: \(M = -\frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right)\), and the tube is slightly longer: \(L = f_o + u_e\) with \(u_e = \frac{f_e D}{D + f_e}\) in magnitude.
3. Terrestrial telescope: a third lens (inverter) of focal length \(f\) is placed between the two, giving an erect image. In normal adjustment the length is \(L = f_o + f_e + 4f\) and the magnifying power is unchanged.
4. Large focal length and large aperture of the objective give higher magnification and brighter images.

Worked examples: lenses of \(f_o = 16\) cm and \(f_e = 2\) cm. In normal adjustment \(M = 8\), the separation is 18 cm, and an object subtending \(0.5^\circ\) has an image subtending \(4^\circ\). If \(M = 9\) and the separation is 20 cm, then \(f_o = 9f_e\) and \(10 f_e = 20\), so \(f_e = 2\) cm and \(f_o = 18\) cm.

Compound Microscope and Telescope Compared