Q35 of 51 Page 345

What is the height of the final image of the tower if it is formed at 25cm?

Given:


Focal length of the objective lens, f0 = 140 cm


Focal length of the eyepiece, fe = 5 cm


Image is formed at a distance, d = 25cm


The magnification of the eyepiece is given by the relation:


m = 1 + d/fe


⇒ m = 1 + 25/5


⇒ m = 6


Height of the final image H, is given by,


H = m × h2 = 6 × 4.7 = 28.2 cm


Hence, the height of the final image of the tower is 28.2 cm.


More from this chapter

All 51 →
35

For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eyepiece?

 35

If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?

 36

A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20mm apart. If the radius of curvature of the large mirror is 220mm and the small mirror is 140mm, where will the final image of an object at infinity be?

 37

Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig. 9.36. A current in the coil produces a deflection of 3.50 of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?