A small telescope has an objective lens of focal length 150 cm and eyepiece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant obj ects in normal adjustment ? 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 ?

Step 1: Given that the focal length of the objective lens is $f_0=150\text{ cm}$ and the focal length of the eyepiece is $f_e=5\text{ cm}$. Step 2: We can calculate the magnifying power of the telescope in normal adjustment as follows: $m=-\frac{f_0}{f_e}=-\frac{150\text{ cm}}{5\text{ cm}}=-30$ Step 3: The negative sign indicates that the image formed is inverted. Step 4: The height of the tower is given as $h=100\text{ m}$ and the distance of the tower is given as $d=3\text{ km}=3000\text{ m}$. Step 5: The angle subtended by the tower at the objective lens is given by: $\alpha =\frac{h}{d}=\frac{h^{\prime }}{f_0}$ where $h^{\prime }$ is the height of the image of the tower formed by the objective lens. Step 6: We can solve for $h^{\prime }$ as follows: $h^{\prime }=\frac{h{f}_0}{d}=\frac{(100\text{ m})(150\text{ cm})}{(3000\text{ m})}=\frac{(100\text{ m})(1.5\text{ m})}{(3000\text{ m})}=0.05\text{ m}=5.0\text{ cm}$ Therefore, the height of the image of the tower formed by the objective lens is $h^{\prime }=5.0\text{ cm}$.