A small telescope has an objective lens of focal length `144 cm` and an eye-piece of focal length `6.0 cm`. What is the magnifying power of the telescope ? What is the separation between the objective and the eye-piece ?
Text Solution
Verified by Experts
Here, `f_(0) = 144 cm, f_(e) 6.0 cm, m = ?, L = ?` In normal adjustment, `m = (-f_(0))/(f_(e)) = (-144)/(6.0) = - 24` `L = f_(0) + f_(e) = 144 + 6.0 = 150.0 cm`
Topper's Solved these Questions
OPTICS
PRADEEP|Exercise Very short answer question|5 Videos
OPTICS
PRADEEP|Exercise very short answer questions|1 Videos
A telescope has an objective of focal length 100 cm and an eye-piece of focal length 5 cm. What is the magnifying power of the telescope when it is in normal adjustment?
A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length 6.0 cm. The magnifying power of the telescope for distinct vision adjustment is:
A small telescope has an objective lens of focal length 140 cm and eye piece of focal length 5.0 cm . What is the magnifying power of telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e. when the image is at infinity) (b) the final image is formed at the least distance of distinct vision (25 cm) .
A telescope has an objective of focal length 100 cm and an eyepiece of focal length 5 cm. What is the magnifying power of the telescope when the final image is formed at the least distance of distinct vision ?
A telescope has an objective of focal length 30 cm and an eyepiece of focal length 3 cm. It is used to focus on an object at a distance of 2 m. Calculate the separation between the objective and the eyepiece for seeing with relaxed eye.
A telescope has an objective of focal length 30 cm and an eye piece of focal length 3.0 cm . It is focussed on a scale distant 2.0 m . For seeing with relaxed aye, calculate the separation between the objective and eye piece.