The value of numerical aperture of the objective lens of a microscope is 1.25. If light of wavelengt

Q: What exactly is Numerical Aperture (NA)?

Numerical Aperture is a dimensionless number that characterizes the range of angles over which the system can accept or emit light. For a microscope, it's defined as $\text{NA} = n \sin\theta$, where $n$ is the refractive index of the medium between the objective lens and the specimen, and $\theta$ is the half-angle of the maximum cone of light that can enter the lens.

Q: Why is there a 0.61 factor in the formula?

The factor $0.61$ comes from the Rayleigh criterion for circular apertures. The first diffraction minimum of a circular aperture occurs at an angle $\theta = 1.22 \frac{\lambda}{D}$. When we derive the limit of resolution for a microscope, this $1.22$ gets divided by $2$, resulting in the $0.61$ factor in the numerator.

Q: How does wavelength affect the resolving power?

Resolving power is inversely proportional to the limit of resolution $d$. Since $d$ is directly proportional to the wavelength $\lambda$, using a shorter wavelength (like blue or ultraviolet light) decreases $d$, which means the microscope can resolve smaller details, thereby increasing its resolving power.

Q: What is the difference between resolving power and limit of resolution?

They are reciprocals of each other. The limit of resolution ($d$ or $\Delta\theta$) is the minimum distance or angle between two objects for them to be seen as separate. Resolving power is the ability to do this, defined mathematically as $\frac{1}{d}$ or $\frac{1}{\Delta\theta}$. A smaller limit of resolution means a higher resolving power.

Q: I always get confused converting Angstroms to micrometers. What's the easiest way?

Think of it in steps through meters. $1\text{ \AA} = 10^{-10}\text{ m}$. And $1\text{ }\mu\text{m} = 10^{-6}\text{ m}$. So, to go from Angstroms to micrometers, you multiply by $10^{-4}$. For example, $5000\text{ \AA} = 5000 \times 10^{-4}\text{ }\mu\text{m} = 0.5\text{ }\mu\text{m}$.

Question

The value of numerical aperture of the objective lens of a microscope is $1.25$. If light of wavelength $5000	ext{ \AA}$ is used, the minimum separation between two points, to be seen as distinct, will be

Options:

(A) $0.24	ext{ }\mu	ext{m}$

(B) $0.38	ext{ }\mu	ext{m}$

(C) $0.12	ext{ }\mu	ext{m}$

(D) $0.48	ext{ }\mu	ext{m}$

Sigma Percentile · Accepted Answer

### The Quest for Clarity: Resolving Power of a Microscope

Have you ever looked through a microscope and struggled to tell if you were looking at one tiny speck or two? That's where the concept of **resolving power** comes in. It's the ability of an optical instrument to produce separate images of two closely spaced objects.

In this problem, we are asked to find the minimum separation between two points so they can be seen as distinct. This minimum separation is also known as the **limit of resolution**.

### The Master Equation

For a microscope, the limit of resolution $d$ is given by a beautiful and precise formula:

$$d = \frac{0.61 \lambda}{	ext{NA}}$$

Here, $\lambda$ is the wavelength of the light being used, and $	ext{NA}$ stands for the **Numerical Aperture** of the objective lens. The Numerical Aperture is a measure of the lens's ability to gather light and resolve fine specimen detail at a fixed object distance.

### Crunching the Numbers

Let's gather our given values:
- Wavelength, $\lambda = 5000	ext{ \AA}$
- Numerical Aperture, $	ext{NA} = 1.25$

First, we must ensure our units are consistent. Let's convert the wavelength from Angstroms to meters:

$$\lambda = 5000 	imes 10^{-10}	ext{ m}$$

Now, we substitute these values into our master equation:

$$d = \frac{0.61 	imes 5000 	imes 10^{-10}}{1.25}$$

Let's simplify the numerator. Multiplying $0.61$ by $5000$ gives us $3050$. So, the equation becomes:

$$d = \frac{3050 	imes 10^{-10}}{1.25}$$

To make the division easier, we can rewrite the numerator by shifting the decimal point:

$$d = \frac{3.05 	imes 10^{-7}}{1.25}$$

### The Final Reveal

Now, for the final calculation. Dividing $3.05$ by $1.25$ yields exactly $2.4$.

$$d = 2.4 	imes 10^{-7}	ext{ m}$$

Looking at the options, they are given in micrometers ($\mu	ext{m}$). Recall that $1	ext{ }\mu	ext{m} = 10^{-6}	ext{ m}$. Let's adjust our decimal point one more time to match the desired units:

$$d = 0.24 	imes 10^{-6}	ext{ m}$$

$$d = 0.24	ext{ }\mu	ext{m}$$

And there we have it! The minimum separation required to see the two points distinctly is **$0.24	ext{ }\mu	ext{m}$**. This means if the points are any closer than this, they will blur into a single, indistinguishable blob.

Animated Solution for Physics - Optics: The value of numerical aperture of the objective lens of a microscope is $1.25$ . If light of wavelength $5000 \overset{˚}{A}$ is used, the minimum separation between two points, to be seen as distinct, will be

Select Answer:

Visualized Solution

$Resolving Power of Microscope$

$Substituting Given Values$

$Simplifying the Expression$

$Final Calculation$

The Sigma Insight: Optical Instruments

The Quest for Clarity

The Master Equation

Crunching the Numbers

The Final Reveal

Similar Questions

Assuming human pupil to have a radius of $0.25 cm$ and a comfortable viewing distance of $25 cm$ , the minimum separation between two objects that human eye can resolve at $500 nm$ wavelength is

Diameter of the objective lens of a telescope is $250 cm$ . For light of wavelength $600 nm$ coming from a distant object, the limit of resolution of the telescope is close to

Two point white dots are $1 mm$ apart on a black paper. They are viewed by eye of pupil diameter $3 mm$ . Approximately, what is the maximum distance at which these dots can be resolved by the eye?\n[Take wavelength of light $= 500 nm$ ]

Calculate the limit of resolution of a telescope objective having a diameter of $200 cm$ , if it has to detect light of wavelength $500 nm$ coming from a star.

The aperture of a telescope is $5 m$ . The separation between the moon and the earth is $4 \times 1 0^{5} km$ . With light of wavelength of $5500 \overset{˚}{A}$ , the minimum separation between objects on the surface of moon, so that they are just resolved, is close to

A compound microscope consists of an objective lens of focal length $1$ cm and an eye piece of focal length $5$ cm with a separation of $10$ cm. The distance between an object and the objective lens, at which the strain on the eye is minimum is $\frac{n}{40}$ cm. The value of $n$ is …… .

Wavelength of light used in an optical instrument are $λ_{1} = 4000 \overset{˚}{A}$ and $λ_{2} = 5000 \overset{˚}{A}$ , then ratio of their respective resolving powers (corresponding to $λ_{1}$ and $λ_{2}$ ) is

In a compound microscope, the magnified virtual image is formed at a distance of 25 cm from the eye-piece. The focal length of its objective lens is 1 cm. If the magnification is 100 and the tube length of the microscope is 20 cm, then the focal length of the eye-piece lens (in cm) is ......... .