Problem 49 - Master Resource Book in JEE Main Physics

Question

The focal length of objective and eyepiece of a microscope are \(1 \mathrm{~cm}\) and \(5 \mathrm{em}\) respectively. If the magnifying power for relaxed eye is 45 , then of the tube is (a) \(9 \mathrm{~cm}\) (b) \(15 \mathrm{~cm}\) (c) \(12 \mathrm{~cm}\) (d) \(6 \mathrm{~cm}\)

Step-by-Step Solution

Answer

The correct answer is not provided, as calculated tube length is 7.5 cm.

Step 1: Understand Given Values

We have an objective lens with a focal length of \( f_o = 1 \text{ cm} \) and an eyepiece with a focal length of \( f_e = 5 \text{ cm} \). The magnifying power \( M \) for relaxed eye is 45.

Step 2: Recall the Magnifying Power Equation

For a microscope, the magnifying power \( M \) when the eye is relaxed is given by: \[ M = \frac{L}{f_o} \times \left(1 + \frac{D}{f_e}\right) \] where \( L \) is the length of the microscope tube and \( D \) is the least distance of distinct vision (usually taken as 25 cm).

Step 3: Substitute Known Values

Plug the known values into the equation: \[ 45 = \frac{L}{1} \times \left(1 + \frac{25}{5}\right) \]. The ratio in the parenthesis is simplified first: \[ 1 + \frac{25}{5} = 1 + 5 = 6 \].

Step 4: Solve for Tube Length \( L \)

Now, substitute back into the equation: \[ 45 = L \times 6 \] and solve for \( L \): \[ L = \frac{45}{6} = 7.5 \text{ cm}. \] Since \( 7.5 \) cm is not provided as an option, let's re-check the possible near options or calculations.

Step 5: Error Review for Final Answer

Upon re-evaluation or alternative calculations methods (if any error is suspected), find the nearest rational choice with provided options. Logical checks or additional multipliers may need consideration to match educational materials if results differ due to interpretation or given task constraints.

Key Concepts

Focal length

The focal length is a crucial concept in optics, especially when dealing with lenses and microscopes. It's defined as the distance between the lens and the point where it converges light to focus. In microscopes, we have two main lenses:

• Objective lens: This lens has a short focal length and provides a high level of magnification. It's typically closer to the specimen being observed. In our exercise, the objective lens has a focal length (\( f_o \)) of \(1 \text{ cm}\).
• Eyepiece lens: Also known as the ocular lens, it has a longer focal length and is used to magnify the image formed by the objective lens. In our case, the eyepiece lens has a focal length (\( f_e \)) of \(5 \text{ cm}\).

Understanding these focal lengths is key to calculating the magnification power and the proper setup of the microscope.

Lens formula

The lens formula is fundamental when working with optical devices. It relates several important variables that describe how lenses function. For a microscope, the magnification power (\( M \)) when the eye is relaxed is expressed by the formula:\[ M = \frac{L}{f_o} \times \left( 1 + \frac{D}{f_e} \right) \]Here,

• \( L \) is the tube length of the microscope.
• \( f_o \) and \( f_e \) represent the focal lengths of the objective and eyepiece lenses respectively.
• \( D \) is the least distance of distinct vision, commonly taken as \(25 \text{ cm}\).

This formula highlights how the magnifying power depends on both the tube length and the focal lengths of the lenses involved. By understanding and applying this equation, one can determine the required settings for a desired level of magnification.

Tube length calculation

Calculating the tube length (\( L \)) is vital in setting up a microscope correctly. The tube length is the distance between the objective lens and the eyepiece. It plays a critical role in achieving the right magnification.In our given exercise, we have:

• Magnifying power (\( M \)) of \(45\).
• Objective lens focal length (\( f_o \)) of \(1 \text{ cm}\).
• Eyepiece lens focal length (\( f_e \)) of \(5 \text{ cm}\).

Using the lens formula,\[ 45 = \frac{L}{1} \times \left(1 + \frac{25}{5}\right) \]Simplifying the expression inside the parentheses gives \(6\). So, \( 45 = L \times 6 \). Solving for the tube length, \( L = \frac{45}{6} = 7.5 \text{ cm} \). Although this result was not among the options provided, it's important to double-check calculations and logical options to ensure consistency with educational objectives.