Problem 83 - Master Resource Book in JEE Main Physics

Question

The magnifying power of an astronomical telescope is 10 and the focal length of its eye-piece is \(20 \mathrm{~cm}\). The focal length of its objective will be (a) \(200 \mathrm{~cm}\) (b) \(2 \mathrm{~cm}\) (c) \(0.5 \mathrm{~cm}\) (d) \(0.5 \times 10^{-2} \mathrm{~cm}\)

Step-by-Step Solution

Answer

The focal length of the objective is 200 cm.

Step 1: Understanding Magnifying Power

The magnifying power of the telescope is given as 10. Magnifying power (M) for an astronomical telescope is defined as the ratio of the focal length of the objective (F_o) to the focal length of the eyepiece (F_e). Thus the formula is \[ M = \frac{F_o}{F_e} \].

Step 2: Substitute the Known Values

We know the magnifying power \( M = 10 \) and the focal length of the eyepiece \( F_e = 20 \text{ cm} \). So, substitute these values into the formula. \[10 = \frac{F_o}{20} \].

Step 3: Solve for Objective Focal Length

To find the focal length of the objective \( F_o \), rearrange the formula to \( F_o = 10 \times 20 \). Calculate \[ F_o = 200 \text{ cm} \].

Key Concepts

Magnifying Power

The concept of magnifying power is crucial when discussing astronomical telescopes. Magnifying power refers to how much larger a telescope can make distant objects appear compared to the naked eye. It is denoted by the letter 'M'.
In an astronomical telescope, the magnifying power is determined by the ratio of the focal length of the objective lens to the focal length of the eyepiece lens. This is represented by the formula: \[ M = \frac{F_o}{F_e} \]Where:

• \( M \) is the magnifying power,
• \( F_o \) is the focal length of the objective lens, and
• \( F_e \) is the focal length of the eyepiece lens.

A higher magnifying power means that the telescope can make objects appear larger, allowing astronomers to observe finer details.

Focal Length

Focal length is a key concept in optics, particularly when discussing lenses and telescopes. It is the distance from the center of a lens or mirror to the focal point, where light rays converge to form a clear image.
Focal length greatly influences the telescope's magnifying power and overall performance. For telescopes, two focal lengths are crucial: the focal length of the objective lens and the focal length of the eyepiece lens.
Understanding how focal length works helps in determining how the telescope magnifies objects. For example, if you double the focal length of the objective lens, you double the magnification, assuming the eyepiece lens' focal length remains constant.

Objective Lens

The objective lens is a vital component of an astronomical telescope. Its primary role is to gather as much light as possible from distant objects and focus it to form a clear image.
This lens determines the telescope's light-gathering power. A larger objective lens can collect more light, which is especially important for viewing faint celestial objects like distant stars and galaxies. The objective lens also significantly affects the magnifying power, as its focal length is a key part of the calculation.
In the given exercise, calculating the objective lens's focal length helps determine the telescope's magnifying power, ensuring the celestial images are not only larger but also clearer.

Eyepiece Lens

The eyepiece lens is another crucial part of an astronomical telescope. It acts as a magnifier for the image formed by the objective lens, allowing the observer to see the magnified image more clearly.
While the objective lens collects light and focuses it into an image, the eyepiece lens takes this image and magnifies it for viewing. The focal length of the eyepiece lens is essential in adjusting the telescope's magnifying power. A shorter focal length for the eyepiece generally means higher magnification.
In the exercise mentioned, the eyepiece lens with a focal length of 20 cm is part of the calculation to find the overall magnifying power of the telescope, demonstrating its role in producing an enlarged view of distant phenomena.