The focal lengths of the objective lens and t | vedclass

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

Question

(D) Given: Focal length of objective,$f_{o} = 30\, cm$.Focal length of eye lens,$f_{e} = 3.0\, cm$.The least distance of distinct vision,$D = 25\, cm$

Vedclass · Accepted Answer

$+8.8$

Vedclass · Answer

(D) Given: Focal length of objective,$f_{o} = 30\, cm$.Focal length of eye lens,$f_{e} = 3.0\, cm$.The least distance of distinct vision,$D = 25\, cm$.For a Galilean telescope,the magnifying power $M$ when the final image is formed at the least distance of distinct vision is given by:$M = \frac{f_{o}}{f_{e}} \left( 1 - \frac{f_{e}}{D} ight)$Substituting the values:$M = \frac{30}{3} \left( 1 - \frac{3}{25} ight)$$M = 10 	imes \left( \frac{25 - 3}{25} ight)$$M = 10 	imes \frac{22}{25}$$M = \frac{220}{25} = 8.8$Since the image is erect,the magnifying power is positive. Thus,$M = +8.8$.

Similar Questions

$A$ planet is observed by an astronomical refracting telescope having an objective of focal length $16 \, m$ and an eye-piece of focal length $2 \, cm$. Which of the following statements is correct?

Magnification produced by an astronomical telescope for normal adjustment is $10$ and the length of the telescope is $1.1 \, m$. The magnification when the image is formed at the least distance of distinct vision $(D = 25 \, cm)$ is

Four lenses of focal length $\pm 15\; cm$ and $\pm 150 \;cm$ are available for making a telescope. To produce the largest magnification,the focal length (in $cm$) of the eyepiece should be

The magnifying power of a telescope with tube length $60 \; cm$ is $5$. What is the focal length of its eyepiece in $cm$?

The magnifying power of a telescope is high if its objective and eyepiece have respectively

Vedclass Test Series

Exam Paper Generator

Online Exam Module