A telescope has an objective lens of focal le | vedclass

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

Question

(A) The height of the image $h'$ formed by the objective lens can be calculated using the magnification formula for a lens: $m = \frac{h'}{h} = \frac{

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The height of the image $h'$ formed by the objective lens can be calculated using the magnification formula for a lens: $m = \frac{h'}{h} = \frac{v}{u}$.Given:Height of the object $h = 50 \ m = 5000 \ cm$.Distance of the object $u = -2 \ km = -2000 \ m = -200000 \ cm$.Focal length of the objective lens $f = 200 \ cm$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v} - \frac{1}{-200000} = \frac{1}{200}$$\frac{1}{v} = \frac{1}{200} - \frac{1}{200000} = \frac{1000 - 1}{200000} = \frac{999}{200000}$$v = \frac{200000}{999} \approx 200.2 \ cm$.Now,$h' = h 	imes \frac{v}{|u|} = 5000 	imes \frac{200000/999}{200000} = \frac{5000}{999} \approx 5.005 \ cm$.Rounding to the nearest integer,the height of the image is $5 \ cm$.

$A$ telescope has an objective lens of focal length $200 \ cm$ and an eyepiece with focal length $2 \ cm$. If this telescope is used to see a $50 \ m$ tall building at a distance of $2 \ km$,what is the height of the image of the building formed by the objective lens in $cm$?

Similar Questions

$A$ line passing through the point of intersection of $x+y=4$ and $x-y=2$ makes an angle $\tan ^{-1}\left(\frac{3}{4}\right)$ with the $x$-axis. It intersects the parabola $y^{2}=4(x-3)$ at points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ respectively. Then, $|x_{1}-x_{2}|$ is equal to

When $FeCl_3$ reacts with $K_4[Fe(CN)_6]$,which of the following compounds is formed?

If $(2,-1,3)$ is the foot of the perpendicular drawn from the origin to the plane,then the equation of the plane is

If $f:[-2,2] \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{\sqrt{1+cx}-\sqrt{1-cx}}{x} & \text{for } -2 \leq x < 0 \\ \frac{x+3}{x+1} & \text{for } 0 \leq x \leq 2 \end{cases}$ and is continuous on $[-2,2]$,then $c$ is equal to

Assertion: The isothermal curves intersect each other at a certain point.
Reason: The isothermal change takes place slowly,so,the isothermal curves have very little slope.

Vedclass Test Series

Exam Paper Generator

Online Exam Module