Problem 69 - Master Resource Book in JEE Main Physics

Question

The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eyepiece is found to be \(20 \mathrm{~cm}\). The focal lengths of the lenses are (a) \(18 \mathrm{~cm}, 2 \mathrm{~cm}\) (b) \(11 \mathrm{~cm}, 9 \mathrm{~cm}\) (c) \(10 \mathrm{~cm}, 10 \mathrm{~cm}\) (d) \(15 \mathrm{~cm}, 5 \mathrm{~cm}\)

Step-by-Step Solution

Answer

The focal lengths are (a) \(18 \text{ cm}, 2 \text{ cm}\).

Step 1: Understand the Formula for Magnifying Power

The magnifying power \( M \) for a telescope with objective lens focal length \( f_o \) and eyepiece focal length \( f_e \) is given by the formula: \( M = \frac{f_o}{f_e} \). We need to find \( f_o \) and \( f_e \) such that \( M = 9 \).

Step 2: Understand Total Length in Telescope

When a telescope is adjusted for parallel rays, the distance between the objective and the eyepiece, \( d \), is the sum of the focal lengths: \( f_o + f_e = 20 \).

Step 3: Set Up the Equations

We now have two equations: \( \frac{f_o}{f_e} = 9 \) and \( f_o + f_e = 20 \). We need to solve these equations simultaneously to find the focal lengths.

Step 4: Solve for One Focal Length in Terms of the Other

From \( \frac{f_o}{f_e} = 9 \), we can express \( f_o \) as \( f_o = 9f_e \). Substitute \( f_o = 9f_e \) in \( f_o + f_e = 20 \), yielding \( 9f_e + f_e = 20 \).

Step 5: Simplify and Solve for \( f_e \)

Combine terms to obtain \( 10f_e = 20 \), leading to \( f_e = 2 \text{ cm} \).

Step 6: Find \( f_o \) Using \( f_e \)

Substitute \( f_e = 2 \text{ cm} \) back into \( f_o = 9f_e \), resulting in \( f_o = 9 \times 2 = 18 \text{ cm} \).

Step 7: Verify the Solution

Check the calculated focal lengths: \( f_o = 18 \text{ cm} \) and \( f_e = 2 \text{ cm} \). They should satisfy both the total length and magnification properties: \( 18 + 2 = 20 \) and \( \frac{18}{2} = 9 \).

Key Concepts

Focal Length of Lenses

The focal length of lenses in a telescope is crucial in determining its magnifying power. In optics, "focal length" represents the distance from the lens to the point where parallel rays of light converge. For a telescope, the focal lengths of its lenses—the objective lens and the eyepiece—work together to magnify distant objects. The objective lens has a longer focal length as it collects light from a distant source. Meanwhile, the eyepiece has a shorter focal length to magnify the image produced by the objective lens. This combination ensures a clear and enlarged view of the object being observed. For this particular exercise, the objective lens’s focal length and the eyepiece’s focal length were calculated using key data: a magnifying power of 9 and a total lens distance of 20 cm. The formulas for these settings help us express these focal lengths in terms of measurable quantities.

Parallel Rays Adjustment

Adjusting a telescope for parallel rays is a method to ensure a clear and focused image. When a telescope is adjusted in this way, the light entering the telescope is assumed to come parallel, which is a typical condition for observing celestial objects. To achieve this, the focal points of the objective lens and the eyepiece coincides. In practice, adjusting for parallel rays involves setting the telescope such that the sum of the focal lengths of the objective and eyepiece equals the distance between them. For instance, in our exercise, this distance was given as 20 cm. It is this setup that ensures the telescope forms an image with rays parallel to the original direction, providing clarity while minimizing strain on the eyes.

Simultaneous Equations in Optics

Simultaneous equations are useful tools in optics to find unknown quantities, such as the focal lengths of lenses. In this exercise, the simultaneous equations assist in linking magnifying power and total distance between lenses to their respective focal lengths.The first equation, derived from the magnification formula \( M = \frac{f_o}{f_e} \), gives a relation between the focal lengths. The second equation, \( f_o + f_e = 20 \), relates to the physical setup of the telescope.By solving these equations together:

• We start by expressing one variable, such as \( f_o \), in terms of the other variable \( f_e \).
• Next, we substitute \( f_o \) back into the second equation to simplify and solve for one unknown, such as \( f_e \).
• This solution is then used to find the other focal length by back-substitution.

This process efficiently solves for both focal lengths, ensuring the given magnification and length conditions. This basic algebraic method simplifies otherwise complex optical problems.