Question

A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the angle formed by the image of the tower is $\theta$, then $\theta$ close to.

• A. ${60}^o$
• B. $1^o$
• C. ${30}^o$
• D. ${15}^o$

Answer 1

The correct option is A ${60}^o$

The tower PQ in figure (19.5) subtends an angle $\alpha$ on the objective. As $u_o$ is very large, the first image P Q' is formed in the focal plane of the objective.

$tan\alpha =\alpha =\frac{P^{\prime }{Q}^{\prime }}{f\circ }$ ..(i)

The final image P"Q" subtends an angle $\beta$ on the eyepiece (and

hence on the eye). We have from the triangle P Q'E

$tan\beta =\beta =\frac{P^{\prime }{Q}^{\prime }}{E{P}^{\prime }}$ ...(ii)

the telescope is set for normal adjustment so that the final image is formed at infinity, the first image P'Q' must be in the focal plane of the eyepiece.Then EP

=$f_e$ . Thus, equation (ii) becomes

$tan\beta =\beta =\frac{f\circ }{f_e}$ ...(iii)

dividing equation (iii) by (i)

$\frac{\beta }{\alpha }=\frac{f\circ }{f_e}$...(iv)

from equation (i) The $\alpha =\frac{P^{\prime }{Q}^{\prime }}{f\circ }$

$P^{\prime }{Q}^{\prime }=m{h}_1$ ...(v) where m is magnification of objective lens

$m=\frac{v}{u}$ where v is position of first image P'Q' and u is position of tower $v=f\circ =150cm,u=1000m$ substituting value of m in equation (v) m

$P^{\prime }{Q}^{\prime }=\frac{.15}{1000}50m=.0075m$

$\alpha =\frac{.0075}{.15}=.05$

value of $\beta$ can be obtained from equation (v)

$tan\beta =\frac{.15}{.05}\times \alpha =30\times .05=1.5rad$

$\beta =56.3$