The Eye, Microscopes and Telescopes: Quick Revision

A brief note on The Eye, Microscopes and Telescopes. Read it before a test. Try each question at the end before you open its answer.

The eye

• Image: real, inverted and smaller, on the retina. The distance from the lens to the retina is fixed. The focal length changes: this is accommodation.
• Normal eye: near point $D = 25\ \text{cm}$ (by convention), far point at infinity.
• Power of a lens: $P = \dfrac{1}{f}$ with $f$ in metre. The unit is the dioptre (D). Convex is positive and concave is negative.
• Short sight: the far point is at $d_f$. A concave lens with $f = -d_f$ and $P = -\dfrac{1}{d_f}$.
• Long sight: the near point is at $d_n$. A convex lens with $P = \dfrac{1}{0.25\ \text{m}} - \dfrac{1}{d_n}$.
• Presbyopia: a bifocal lens. Astigmatism: a cylindrical lens.
• A lens at a distance $s$ from the eye: measure the distances from the lens. For short sight, $f = -(d_f - s)$.

Microscopes

• Magnifying power: $m = \dfrac{\beta}{\alpha}$, a ratio of angles, with $D = 25\ \text{cm}$.
• Simple microscope: $m = 1 + \dfrac{D}{f}$ (image at the near point) and $m = \dfrac{D}{f}$ (image at infinity).
• Compound microscope: $M = m_om_e$. The objective has a short focal length and forms a real, inverted, magnified image.
• Image at infinity: $|M| = \dfrac{L}{f_o}\cdot\dfrac{D}{f_e}$. $L$ is the distance between the focal points. The lenses are $f_o + L + f_e$ apart.
• Limit of resolution: $d_{\min} = \dfrac{1.22\,\lambda}{2n\sin\beta}$. Shorter wavelength and oil immersion make it smaller.

Telescopes

• Astronomical, normal adjustment: $|m| = \dfrac{f_o}{f_e}$ and length $f_o + f_e$. The image is inverted.
• Final image at the near point: $|m| = \dfrac{f_o}{f_e}\left(1 + \dfrac{f_e}{D}\right)$.
• Large aperture: it gathers light in proportion to $D^2$ and resolves $\theta = \dfrac{1.22\,\lambda}{D}$.
• Reflecting telescopes: no chromatic aberration, the mirror is supported from behind, and a parabolic mirror removes spherical aberration. Newtonian: a flat mirror at $45^\circ$. Cassegrain: a convex secondary mirror and a hole in the primary.
• Terrestrial: erect image, length $f_o + 4f + f_e$. Galilean: concave eyepiece, erect image, length $f_o - |f_e|$.

A worked example

A woman has a near point of $75\ \text{cm}$. Find the power of her reading glasses, at the eye.

1. The book. It is at $u = -25\ \text{cm}$, and the image must be at $v = -75\ \text{cm}$.
2. The power. $P = \dfrac{1}{0.25\ \text{m}} - \dfrac{1}{0.75\ \text{m}} = 4.00\ \text{D} - 1.33\ \text{D} = +2.67\ \text{D}$.
3. The focal length. $f = \dfrac{1}{2.67\ \text{D}} = 0.375\ \text{m}$.

Practise with new numbers

Check yourself

1. A microscope has $f_o = 0.50\ \text{cm}$, $f_e = 2.5\ \text{cm}$ and $L = 18\ \text{cm}$. What is $|M|$ with the final image at infinity?

2. A telescope in normal adjustment has $|m| = 18$ and a length of $76\ \text{cm}$. What is $f_o$?

3. What is the smallest angle that a telescope with a mirror of diameter $2.0\ \text{m}$ can separate, for light of $500\ \text{nm}$?

Sources

• NCERT Physics Part II (Class 12), Chapter 9, the sections on the human eye and the optical instruments.
• OpenStax University Physics Volume 3, Chapter 2, the sections on the microscope and the telescope.
• OpenStax College Physics 2e, Chapter 26, Vision and Optical Instruments.