Optics

In normal adjustment of an astronomical telescope (final image at infinity):

• Tube length $L$ is the sum of focal lengths of objective and eyepiece:
  $$f_o = 10\,\text{m} = 1000\,\text{cm}$$
  $$f_e = 10\,\text{cm}$$
  $$L = f_o + f_e = 1000\,\text{cm} + 10\,\text{cm} = 1010\,\text{cm}$$
• The magnifying power (magnification) in normal adjustment is:
  $$m = \frac{f_o}{f_e} = \frac{1000\,\text{cm}}{10\,\text{cm}} = 100$$

For total internal reflection (TIR) to occur, two essential conditions must be met:

• Light must travel from an optically denser medium (higher refractive index) to an optically rarer medium (lower refractive index).
• The angle of incidence in the denser medium must exceed the critical angle ($i \gt i_c$).
  Comparing the refractive indices:
  • Glass ($\mu \approx 1.5$)

• Water ($\mu \approx 1.33$)
• Air ($\mu \approx 1.0$)

Light travels from a denser medium to a rarer medium ONLY when moving from Glass to Water (or Glass to Air, Water to Air).

In options 1, 2, and 3, light travels from a rarer to a denser medium, where TIR is impossible.

When the final image is formed at the near point (least distance of distinct vision $D = 25\,\text{cm}$), the magnifying power of an astronomical telescope is:

$$m = \frac{f_o}{f_e} \left(1 + \frac{f_e}{D}\right)$$

Given:

• $f_o = 50\,\text{cm}$
• $f_e = 2\,\text{cm}$
• $D = 25\,\text{cm}$

Calculating magnification $m$:

$$m = \frac{50}{2} \left(1 + \frac{2}{25}\right) = 25 \times \left(\frac{27}{25}\right) = 27$$

The angular magnification is also defined as the ratio of the angular size of the image ($\beta$) to the angular size of the object ($\alpha$):

$$m = \frac{\beta}{\alpha} \implies \beta = m \times \alpha$$

Given $\alpha = \left(\frac{1}{2}\right)^\circ$:

$$\beta = 27 \times \left(\frac{1}{2}\right)^\circ = 13.5^\circ$$

By Lens Maker's Formula for a lens of refractive index $\mu_L$ immersed in a medium of refractive index $\mu_M$:

$$\frac{1}{f} = \left(\frac{\mu_L}{\mu_M} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

For the biconvex cavity geometry with $R_1 = +10\,\text{cm}$ and $R_2 = -10\,\text{cm}$:

$$\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = \frac{1}{10} - \left(-\frac{1}{10}\right) = \frac{2}{10} = \frac{1}{5}\,\text{cm}^{-1}$$

Case 1 (Air cavity in glass: $\mu_L = 1$, $\mu_M = 1.5$):

$$\frac{1}{f_1} = \left(\frac{1}{1.5} - 1\right) \left(\frac{1}{5}\right) = \left(\frac{2}{3} - 1\right) \left(\frac{1}{5}\right) = -\frac{1}{3} \times \frac{1}{5} = -\frac{1}{15}\,\text{cm}^{-1}$$ $$f_1 = -15\,\text{cm}$$

Case 2 (Liquid lens in glass: $\mu_L = 2$, $\mu_M = 1.5$):

$$\frac{1}{f_2} = \left(\frac{2}{1.5} - 1\right) \left(\frac{1}{5}\right) = \left(\frac{4}{3} - 1\right) \left(\frac{1}{5}\right) = +\frac{1}{3} \times \frac{1}{5} = +\frac{1}{15}\,\text{cm}^{-1}$$ $$f_2 = +15\,\text{cm}$$