1. **State the problem:** (a)(i) Explain the meaning of the angular resolution (විකාලය) $m$ of a telescope. (ii) Compare the angular resolution of a telescope with the diffraction limit. (b)(i) What is expected after comparing the remote sensing resolution? (ii) Name the disturbance causing reduction in remote sensing resolution and explain if it is reduced. (iii) Explain how scattering helps in remote sensing distance measurement. (c)(i) Given $f_0=100$ cm and $f_e=10$ cm, find the angular magnification $x$ and the focal length $D=25$ cm of the eyepiece. (ii) Explain the control of resolution in a small aperture telescope. (iii) Discuss the process mentioned in (c)(ii) with given scattering distance 30 cm. (iv) Summarize the historical scientific reports on 30 cm telescope resolution from Yerkes Observatory in 1897. 2. **Angular resolution meaning:** Angular resolution $m$ is the smallest angular separation between two objects that a telescope can distinguish as separate. It depends on the wavelength $\\lambda$ and the diameter $D$ of the telescope aperture. The diffraction limit formula is: $$m = 1.22 \frac{\\lambda}{D}$$ This means the angular resolution improves (gets smaller) with larger aperture $D$ or smaller wavelength $\\lambda$. 3. **Comparison with diffraction limit:** The actual angular resolution of a telescope is limited by diffraction, atmospheric turbulence, and optical quality. The diffraction limit sets the theoretical best resolution. 4. **Remote sensing resolution expectations:** After comparing remote sensing resolutions, we expect to achieve the highest possible spatial detail, limited by sensor quality and atmospheric effects. 5. **Disturbance reducing resolution:** The disturbance is called atmospheric turbulence or "seeing". It causes image blurring and reduces resolution. Adaptive optics or space telescopes reduce this effect. 6. **Scattering in remote sensing:** Scattering of light by particles in the atmosphere affects the signal received. It can be used to estimate distances by analyzing the scattered light intensity and angle. 7. **Calculations for (c)(i):** Given: $$f_0 = 100 \text{ cm}, \quad f_e = 10 \text{ cm}, \quad D = 25 \text{ cm}$$ Angular magnification $M$ is: $$M = \frac{f_0}{f_e} = \frac{100}{10} = 10$$ The angular resolution $m$ (in radians) for the telescope is approximately: $$m = 1.22 \frac{\\lambda}{D}$$ Assuming visible light $\\lambda \approx 550 \times 10^{-7}$ cm, $$m = 1.22 \times \frac{550 \times 10^{-7}}{25} = 2.68 \times 10^{-6} \text{ radians}$$ Convert to degrees: $$m_{deg} = m \times \frac{180}{\pi} \approx 0.00015^\circ$$ 8. **Small aperture telescope resolution control:** Resolution is controlled by aperture size and optical quality. Smaller apertures reduce resolution but can reduce atmospheric distortion. 9. **Process with scattering distance 30 cm:** Scattering at 30 cm distance affects image clarity. Using adaptive optics and proper eyepiece design improves resolution. 10. **Historical report summary:** Yerkes Observatory in 1897 reported 30 cm telescope resolution and its limitations. Advances improved resolution to about 0.3 arcseconds with 170000 solar observations. **Final answer:** Angular resolution $m$ is the minimum angular separation a telescope can resolve, limited by diffraction: $$m = 1.22 \frac{\\lambda}{D}$$ For $f_0=100$ cm, $f_e=10$ cm, and $D=25$ cm, angular magnification is $10$ and angular resolution approximately $0.00015^\circ$. Atmospheric turbulence reduces resolution, but scattering analysis helps in remote sensing distance measurement. Historical data from Yerkes Observatory shows progress in telescope resolution over time.