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 a normal setting, the angle formed by the image of the tower is $$\theta$$, then $$\theta$$ is close to

To solve this problem, we need to find the angle θ formed by the image of a tower when observed through a telescope in normal adjustment. The telescope has an objective lens with a focal length of 150 cm and an eyepiece with a focal length of 5 cm. The tower is 50 m tall and located at a distance of 1 km. We will proceed step by step, explaining each part clearly.

First, recall that in a telescope set to normal adjustment, the final image is formed at infinity. This occurs when the distance between the objective lens and the eyepiece is equal to the sum of their focal lengths. The focal length of the objective lens is 150 cm, which is 1.5 m (since 1 m = 100 cm, so 150 cm = 150 / 100 = 1.5 m). The focal length of the eyepiece is 5 cm, which is 0.05 m. Therefore, the separation between the lenses is $$1.5 \text{m} + 0.05 \text{m} = 1.55 \text{m}$$.

Next, we need to find the angle subtended by the object (the tower) at the unaided eye. The tower has a height of 50 m and is at a distance of 1 km, which is 1000 m. For small angles, the angle subtended by the object, denoted as θ_o, is approximately equal to the ratio of the height to the distance, in radians. So, $$\theta_o = \frac{\text{height}}{\text{distance}} = \frac{50 \text{m}}{1000 \text{m}} = 0.05 \text{radians}$$. This approximation is valid because the angle is small, and we can use $$\tan \theta_o \approx \theta_o$$ when θ_o is in radians.

Now, we need the angular magnification of the telescope. In normal adjustment, the angular magnification M is given by the ratio of the focal length of the objective lens to the focal length of the eyepiece. Using the focal lengths in consistent units (meters or centimeters), we can use centimeters: focal length of objective $$f_o = 150 \text{cm}$$, focal length of eyepiece $$f_e = 5 \text{cm}$$. Thus, $$M = \frac{f_o}{f_e} = \frac{150}{5} = 30$$. Alternatively, using meters: $$f_o = 1.5 \text{m}$$, $$f_e = 0.05 \text{m}$$, so $$M = \frac{1.5}{0.05} = 30$$. The magnification is dimensionless.

The angular magnification M is defined as the ratio of the angle subtended by the image (θ) to the angle subtended by the object (θ_o). Therefore, $$M = \frac{\theta}{\theta_o}$$. Rearranging this, we find the angle θ as $$\theta = M \times \theta_o$$. Substituting the values, $$\theta = 30 \times 0.05 = 1.5 \text{radians}$$.

Since the options are given in degrees, we need to convert θ from radians to degrees. The conversion factor is that 1 radian equals $$\frac{180}{\pi}$$ degrees, where $$\pi \approx 3.1416$$. So, $$1 \text{radian} = \frac{180}{3.1416} \approx 57.2958 \text{degrees}$$. Therefore, $$\theta = 1.5 \times 57.2958 \approx 85.9437 \text{degrees}$$. This is approximately 86 degrees.

Hence, the angle θ formed by the image of the tower is close to 86 degrees.

So, the answer is Option D.