Question

# A telescope has an objective of focal length 30 cm and an eyepiece of focal length 3.0 cm. It is focussed on a scale present at a distance of 2.0 m. For seeing with relaxed eye, calculate the separation between the objective and the eyepiece.

A
38.3 cm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
42.3 cm
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
35.3 cm
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
49 cm
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is A 38.3 cmFor seeing with relaxed eye, the final image should be formed at infinity. For this, the image formed by the objective should fall at the focus of the eyepiece. Let v0 be the distance of the image formed by the objective. Then, from the lens formula, 1v0−1u0=1f0 Here u0=−2.0 m=−200 cm,f0=30 cm ⇒1v0−1−200=130 ⇒1v0=130−1200=20−3600=17600 ⇒v0=60017=35.3 cm ∴ distance between the objective and the eyepiece, d=v0+fe=35.3+3.0=38.3 cm Hence, option (A) is correct. Why this question? Key Note:For seeing with relaxed eye, the final image should be formed at infinity. For this, the image formed by the objective should fall at the focus of the eyepiece.

Suggest Corrections
2
Join BYJU'S Learning Program
Select...
Related Videos
Lateral Magnification
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program
Select...