A refracting telescope has an objective lens of focal length $120,$cm and an eyepiece lens of focal length $15,$cm. The telescope is adjusted so that the final image is formed at the near point of a relaxed eye ($D=25,$cm). Calculate the angular magnification $M$ of the telescope in this relaxed‑eye configuration.

Step-by-step solution

1. Use the relaxed‑eye formula for telescopes: $M=\frac{f_o}{f_e}\left(1-\frac{f_e}{D}\right)$.
2. Substitute $f_o=120,$cm, $f_e=15,$cm and $D=25,$cm.
3. Compute $1-\frac{f_e}{D}=1-\frac{15}{25}=0.4$.
4. Multiply $\frac{f_o}{f_e}=\frac{120}{15}=8$ by $0.4$ to get $M=3.2$. Wait – we must use the correct relaxed‑eye magnification: $M=\frac{f_o}{f_e}\cdot \frac{D}{f_e+D}$? No, for relaxed eye with final image at near point, formula is $M=\frac{f_o}{f_e}\left(1-\frac{f_e}{D}\right)$. Re‑compute: $1-\frac{15}{25}=0.4$. Thus $M=8\times 0.4=3.2$. But check: Standard relaxed‑eye angular magnification with final image at near point is $M=\frac{f_o}{f_e}\left(1-\frac{f_e}{D}\right)$. So the correct answer is $3.2$. We made mistake earlier: we incorrectly wrote $M=8$. Correcting: $M=3.2$. Final answer: $3.2$. (Note: The earlier answer of 8 corresponds to viewing at infinity, not near point).

Concept tested: Angular magnification of telescope (near‑point adjustment)

Published 22 September 2026 · How this solution was prepared and checked