In an astronomical telescope,the focal length | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For an astronomical telescope,the magnifying power $(m)$ for a normal eye (final image at infinity) is given by the formula:$m = -\frac{f_o}{f_e}$

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(A) For an astronomical telescope,the magnifying power $(m)$ for a normal eye (final image at infinity) is given by the formula:$m = -\frac{f_o}{f_e}$Given:Focal length of the objective lens,$f_o = 100 \, cm$Focal length of the eyepiece,$f_e = 2 \, cm$Substituting the values:$m = -\frac{100}{2} = -50$The magnitude of the magnifying power is $50$. The negative sign indicates that the final image is inverted.

In an astronomical telescope,the focal length of the objective lens is $100 \, cm$ and of the eyepiece is $2 \, cm$. The magnifying power of the telescope for the normal eye is:

Similar Questions

To produce a magnified erect image of a far object,along with a convex lens,we will require:

The focal length of the objective and eyepiece of a telescope are respectively $200 \, cm$ and $5 \, cm$. The maximum magnifying power of the telescope will be

Which are two main points important for an astronomical telescope?

If the focal length of the objective lens is increased,then:

The angular magnification of an astronomical telescope is $10$. The final image is formed at infinity. The length of the telescope is $44 \, cm$. The focal length of the objective lens is .......... $cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module