Question

In a microscope of tube length $10\,\text{cm}$ two convex lenses are arranged with focal lengths $2\,\text{cm}$ and $5\,\text{cm}$. Total magnification obtained with this system for normal adjustment is $(5)^k$. The value of $k$ is ___.

• A. $4$
• B. $5$
• C. $3.5$
• D. $2$

Hint:

For microscopes, objective lens mainly increases linear magnification while eyepiece acts as a magnifier.

Answer 1

The Correct Option is C

To solve this problem, we need to calculate the total magnification of a microscope's optical system comprising two convex lenses with given focal lengths. Let's break down the solution step-by-step:

Given:

• Tube length of the microscope, \(L = 10\,\text{cm}\)
• Focal length of the objective lens, \(f_o = 2\,\text{cm}\)
• Focal length of the eyepiece lens, \(f_e = 5\,\text{cm}\)

The total magnification \(M\) for a microscope is given by the formula:

\(M = \frac{L}{f_o}\cdot\left(1 + \frac{D}{f_e}\right)\)

Where:

• \(D\) is the least distance of distinct vision, typically \(25\,\text{cm}\).

Substituting the given values into the formula:

\(M = \frac{10}{2}\cdot\left(1 + \frac{25}{5}\right)\)

Calculating step-by-step:

1. Calculate the magnification due to the objective lens: \(\frac{L}{f_o} = \frac{10}{2} = 5\).
2. Calculate the magnification produced by the eyepiece: \(1 + \frac{D}{f_e} = 1 + \frac{25}{5} = 1 + 5 = 6\).
3. Total magnification is then: \(M = 5 \times 6 = 30\).

Given that the total magnification is expressed as \((5)^k\), we equate and solve for \(k\):

\((5)^k = 30\)

Taking logarithm base 5 on both sides:

\(k = \log_{5} 30\)

Expressing \(30\) in terms of base \(5\):

\(30 = 5^{3.5}\)

Therefore, \(k = 3.5\).

Hence, the value of \(k\) is 3.5.

This corresponds to the correct answer: $3.5$.