In an astronomical telescope in normal adjust | vedclass

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

Question

(A) In normal adjustment,the distance between the objective lens and the eyepiece is $f_0 + f_e$. Since the line of length $L$ is on the objective len

Vedclass · Accepted Answer

$\frac{L}{I}$

Vedclass · Answer

(A) In normal adjustment,the distance between the objective lens and the eyepiece is $f_0 + f_e$. Since the line of length $L$ is on the objective lens,its distance from the eyepiece is $u = -(f_0 + f_e)$. The magnification $m$ produced by a lens is given by $m = \frac{f}{f + u}$. For the eyepiece,the magnification of the image of length $I$ is $\frac{I}{L} = \frac{f_e}{f_e + u}$. Substituting $u = -(f_0 + f_e)$,we get: $\frac{I}{L} = \frac{f_e}{f_e - (f_0 + f_e)} = \frac{f_e}{-f_0}$. Since the magnification of the telescope is $M = -\frac{f_0}{f_e}$,we have $\frac{I}{L} = \frac{1}{M}$. Therefore,the magnification $M = \frac{L}{I}$.

In an astronomical telescope in normal adjustment,a straight black line of length $L$ is drawn on the inside part of the objective lens. The eyepiece forms a real image of this line. The length of this image is $I$. The magnification of the telescope is:

Similar Questions

Which of the following has the smallest size?

The equation of the hyperbola which passes through the point $(2,3)$ and has the asymptotes $4x+3y-7=0$ and $x-2y-1=0$ is

The density $($in $g cm ^{-3} )$ of the metal which forms a cubic close packed $(ccp)$ lattice with an axial distance $($edge length$)$ equal to $400 \ pm$ is $....$
Use: Atomic mass of metal $=105.6 \ amu$ and Avogadro's constant $=6 \times 10^{23} mol^{-1}$

The sum of the fourth powers of the roots of the equation $x^3+x+1=0$ is

In short-day plants,flowering is induced by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module