Problem 28 - Master Resource Book in JEE Main Physics

Question

A telescope has an objective of focal length \(50 \mathrm{~cm}\) and an eyepiece of focal length \(5 \mathrm{~cm}\). The least distance of distinct vision is \(25 \mathrm{~cm}\). The telescope is focussed for distinct vision on a scale \(200 \mathrm{~cm}\) away. The separation between the objective and eyepiece is (a) \(74 \mathrm{~cm}\) (b) \(75 \mathrm{~cm}\) (c) \(60 \mathrm{~cm}\) (d) \(71 \mathrm{~cm}\)

Step-by-Step Solution

Answer

The separation is approximately 71 cm (option d).

Step 1: Understand Telescope Configuration

The telescope has an objective lens with a focal length \( f_o = 50 \text{ cm} \) and an eyepiece with a focal length \( f_e = 5 \text{ cm} \). The telescope is adjusted for distinct vision on an object 200 cm away. We need to determine the separation \( d \) between the objective and the eyepiece lens.

Step 2: Use the Lens Formula for Objective Lens

For the objective, the lens formula \( \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \) can be used, where \( f_o = 50 \text{ cm} \), \( u_o = -200 \text{ cm} \) (since the object distance is real and negative), and \( v_o \) is the image distance. Substitute known values: \[ \frac{1}{50} = \frac{1}{v_o} + \frac{1}{200} \].

Step 3: Solve for the Image Distance of Objective

Rearrange the equation to solve for \( v_o \): \( \frac{1}{v_o} = \frac{1}{50} - \frac{1}{200} = \frac{4}{200} - \frac{1}{200} = \frac{3}{200} \). Therefore, \( v_o = \frac{200}{3} \approx 66.67 \text{ cm} \).

Step 4: Apply the Final Adjustment for Distinct Vision

Using the condition of least distance of distinct vision (\( D = 25 \text{ cm} \)), the telescope is set up such that the image formed by the objective is at a focal point for the eyepiece to produce distinct vision. Thus, the virtual image distance \( v_e = D = 25 \text{ cm} \).

Step 5: Use the Lens Formula for Eyepiece

The eyepiece formula, similar to the objective's, is \( \frac{1}{f_e} = \frac{1}{v_e} - \frac{1}{u_e} \). Substitute values: \( \frac{1}{5} = \frac{1}{25} - \frac{1}{u_e} \).

Step 6: Solve for the Object Distance of Eyepiece

Solve for \( u_e \): \( \frac{1}{u_e} = \frac{1}{25} - \frac{1}{5} = \frac{1}{25} - \frac{5}{25} = -\frac{4}{25} \). Hence, \( u_e = -\frac{25}{4} = -6.25 \text{ cm} \).

Step 7: Calculate the Separation Between Lenses

The separation \( d \) between the lenses is the sum of \( v_o \) and \( |u_e| \) (absolute value, since it is a virtual image): \( d = 66.67 + 6.25 = 72.92 \text{ cm} \). The closest option is \( 71 \text{ cm} \).

Key Concepts

Objective Lens

The objective lens is a crucial component of a telescope, responsible for gathering light from distant objects and creating an image. In a telescope setup, the objective lens typically has a longer focal length compared to the eyepiece lens. This is because it is designed to capture as much light as possible from faraway objects.

The focal length of a lens is the distance from the lens to its focus, where parallel rays converge. In the exercise example, the objective lens has a focal length of 50 cm. Its primary role is to form a real and inverted image of the distant object. The size and brightness of this image largely depend on the lens's diameter and focal length.

• Longer focal lengths allow for more light collection.
• A larger diameter leads to better resolution and brighter images.

This initial image formed by the objective is used and further magnified by the eyepiece lens to give the viewer a detailed look at celestial or distant objects.

Eyepiece Lens

The eyepiece lens in a telescope is tasked with magnifying the image produced by the objective lens. Compared to the objective lens, the eyepiece has a shorter focal length. This shorter focal length is intentionally chosen as it assists in magnifying the initial image created by the objective lens.

In our example, the eyepiece has a focal length of 5 cm. By focusing on the image made by the objective lens, the eyepiece allows the viewer to see a magnified version of it. This is where the actual viewing happens, providing a larger and more detailed view of the faraway object.

• Shorter focal lengths lead to higher magnification.
• The eyepiece's position is adjustable for a comfortable viewing experience.

Through careful adjustment, the eyepiece makes sure that the formed image is within the least distance of distinct vision, optimizing clarity and detail perception.

Lens Formula

The lens formula is a fundamental principle in optics, enabling us to relate the object distance, the image distance, and the focal length of a lens. It is expressed as: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)

Here, \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. This formula is instrumental in handling both the objective and eyepiece lenses in our telescope example.

For the objective lens, the image distance \( v_o \) is calculated as the distance where the initially captured real image is formed. Plugging data into the formula, we found the image distance to be approximately 66.67 cm.

Similarly, the eyepiece uses the formula to establish its virtual object distance. The calculations reveal how the eyepiece focuses on refining the image for optimal viewing, thereby determining the exact positioning of both lenses in the telescope.

• The lens formula applies universally to converging and diverging lenses.
• It provides a systematic route to design effective optical instruments.

Mastering this concept is crucial for anyone engaging in optics or handling vision-related devices.

Least Distance of Distinct Vision

The least distance of distinct vision, often denoted as \(D\), refers to the closest distance at which the eye can comfortably focus on and distinguish an object. In most scenarios, this is accepted to be around 25 cm for a healthy human eye. Adjustments in optical devices like telescopes and microscopes often consider this factor to provide a clear and comfortable viewing experience.

In our telescope exercise, the least distance of distinct vision plays an essential role when adjusting the eyepiece lens. By setting the virtual image distance \(v_e\) equivalent to this least distance (25 cm in this example), viewers can see a detailed and focused image without straining the eyes unnecessarily.

• This distance ensures clarity and precision in viewing experiences.
• It is a part of standard setups in optical devices.

Understanding and using the least distance of distinct vision is vital for creating optimal and user-friendly optical arrangements in telescopes and other vision-aiding devices.