Login/Sign Up
Home » In an astronomical telescope in normal adjustment a straight black line of the length L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is 1 . The magnification of the telescope is: (1) (2) (3) (4)
In an astronomical telescope in normal adjustment a straight black line of the length L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is 1 . The magnification of the telescope is: (1) \frac{L}{I} (2) \frac{\mathrm{L}}{1}+1 (3) \frac{\mathrm{L}}{\mathrm{I}}-1 (4) \frac{\mathrm{L}+1}{\mathrm{~L}-\mathrm{I}}
Physics | Physics
In an astronomical telescope in normal adjustment a straight black line of the length L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is 1 . The magnification of the telescope is: (1) \frac{L}{I} (2) \frac{\mathrm{L}}{1}+1 (3) \frac{\mathrm{L}}{\mathrm{I}}-1 (4) \frac{\mathrm{L}+1}{\mathrm{~L}-\mathrm{I}}

\begin{array}{l} \mathrm{m}=\frac{f}{\mathrm{f}+\mathrm{u}} \\ -\frac{\mathrm{I}}{\mathrm{L}}=\frac{\mathrm{f}_{\mathrm{e}}}{\mathrm{f}_{\mathrm{e}}+\left(-\left(\mathrm{f}_{0}+\mathrm{f}_{\mathrm{e}}\right)\right.} \\ \Rightarrow \frac{1}{\mathrm{~L}}=\frac{\mathrm{f}_{\mathrm{e}}}{\mathrm{f}_{0}} \\ \mathrm{~m} \cdot \mathrm{p} .=\frac{\mathrm{f}_{0}}{\mathrm{f}_{\mathrm{e}}}=\frac{\mathrm{L}}{\mathrm{I}} \end{array}

i

More Material