Question

The magnification produced by an astronomical telescope for normal adjustment is 10 and the length of the telescope is 1.1m. The magnification, when the image is formed at least distance of distinct vision is

• A. 6
• B. 14
• C. 16
• D. 18

Answer 1

Answer: (B)

14

Answer 2

Magnification $\frac{{{f}_{o}}}{{{f}_{e}}}=10$ or ${{f}_{o}}=10{{f}_{e}}$ Given ${{f}_{e}}+{{f}_{o}}=1.1\,m$ ${{f}_{e}}+10{{f}_{e}}=1.1\times 100\,cm$ $11{{f}_{e}}=110$ ${{f}_{e}}=10$ Magnification at least distance of distinct vision ${{M}_{b}}=\frac{{{f}_{o}}}{{{f}_{e}}}\left( 1+\frac{{{f}_{e}}}{D} \right)$ $=10\left( 1+\frac{10}{25} \right)=10\left( \frac{35}{25} \right)=14$