Question

The small telescope has a focal length of $150cm$ and an eyepiece of focal length 5 cm. it is used to view a $100m$ tall tower $3km$ away. What is the height of the image of the tower formed by the objective?

Answer 1

Magnifying power m = $\frac{-f_0}{f}(1+\frac{f_e}{D})$

$f_0$ = focal length of objective = 150 cm

$f_e$ = focal length of eyepiece = 3 cm

D= least distance of ditace vision = 25 cm

$m=\frac{-150}{5}\times (1+\frac{5}{25})=-36$

We know m = $\frac{\beta }{\alpha }$

$m=\frac{tan\beta }{tan\alpha }$

$tan\alpha =\frac{Heightofobject}{Distanceofobjectfromobjective}=\frac{100}{3000}=\frac{1}{30}$

$m=\frac{tan\beta }{\frac{1}{30}}$

$tan\beta =\frac{-36}{30}$

$tan\beta =\frac{Heightofimage}{distanceofimageformation}=\frac{H^{\prime }}{D}$

$H^{\prime }=\frac{-36}{30}\times D=\frac{-36}{30}\times 25=-30cm$

We get a inverted image