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.

$$tan \alpha=\alpha=\dfrac{P'Q'}{f\circ{}}$$ ..(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

$$tan \beta=\beta= \dfrac{P'Q'}{EP'}$$ ...(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

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

dividing equation (iii) by (i)

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

from equation (i) The $$ \alpha =\dfrac{P'Q'}{f\circ{}} $$

$$P'Q'= m h_1$$ ...(v) where m is magnification of objective lens

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

$$P'Q'= \dfrac{.15}{1000} 50m=.0075 m$$

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

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

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

$$ \beta = 56.3$$