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 $\alpha = \left(\frac{1}{2}\right)^\circ$:
$$\beta = 27 \times \left(\frac{1}{2}\right)^\circ = 13.5^\circ$$जब अंतिम प्रतिबिंब स्पष्ट दृष्टि की न्यूनतम दूरी ($D = 25\,\text{cm}$) पर बनता है, तो खगोलीय दूरदर्शक की आवर्धन क्षमता निम्न होती है:
$$m = \frac{f_o}{f_e} \left(1 + \frac{f_e}{D}\right)$$दिए गए मान:
- $f_o = 50\,\text{cm}$
- $f_e = 2\,\text{cm}$
- $D = 25\,\text{cm}$
आवर्धन $m$ की गणना:
$$m = \frac{50}{2} \left(1 + \frac{2}{25}\right) = 25 \times \frac{27}{25} = 27$$कोणीय आवर्धन का सूत्र:
$$m = \frac{\beta}{\alpha} \implies \beta = m \times \alpha$$यहाँ वस्तु (चंद्रमा) का कोणीय व्यास $\alpha = \left(\frac{1}{2}\right)^\circ$ है:
$$\beta = 27 \times 0.5^\circ = 13.5^\circ$$