Angular width of central maximum in the Fraunhofer diffraction pattern of a slit is measured. The sl

Q: Why do we use $\sin \theta \approx \theta$ here?

In Fraunhofer diffraction, the angles involved are typically very small because the wavelength of light is much smaller than the slit width. For small angles expressed in radians, the sine of the angle is approximately equal to the angle itself.

Q: Does the slit width $b$ change when immersed in the liquid?

No, the physical dimensions of the apparatus, including the slit width $b$ and the distance to the screen $D$, remain completely unaffected by the surrounding medium. Only the optical properties of the light change.

Q: Why does the wavelength decrease in a denser medium?

The speed of light decreases in a denser medium ($v = c/\mu$), but the frequency of the light source remains constant. Since $v = f\lambda$, a decrease in speed must result in a proportional decrease in wavelength.

Q: What is the difference between angular width and linear width?

Angular width ($2\theta$) measures the spread of the central maximum in terms of angle and is independent of the screen distance $D$. Linear width ($2y = 2D\theta$) measures the physical spread on the screen and increases as the screen is moved further away.

Q: Can the angular width ever increase when immersed in a liquid?

No, because the refractive index $\mu$ of any liquid is always greater than 1. Therefore, the wavelength in the liquid ($\lambda/\mu$) will always be less than in air, leading to a narrower angular width.

Question

Angular width of central maximum in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength 6000 \AA. When the slit is illuminated by light of another wavelength, the angular width decreases by 30%. Calculate the wavelength of this light. The same decrease in the angular width of central maximum is obtained when the original apparatus is immersed in a liquid. Find refractive index of the liquid.

Sigma Percentile · Accepted Answer

The phenomenon of Fraunhofer diffraction provides a beautiful demonstration of the wave nature of light. When light passes through a narrow slit, it spreads out, creating a pattern of bright and dark fringes on a distant screen.

The most prominent feature of this pattern is the central maximum, which is flanked by the first minima on either side. Understanding the geometry of this central maximum is the key to solving this problem.

### The Geometry of the Central Maximum

Let's consider a slit of width $b$ illuminated by light of wavelength $\lambda$. The light diffracts and forms a pattern on a screen placed at a distance $D$.

The angular position $	heta$ of the first minimum is determined by the condition for destructive interference. This condition is given by the equation:

$$b \sin 	heta = \lambda$$

Because the wavelength of light is typically much smaller than the slit width, the angle $	heta$ is very small. For small angles, we can use the approximation $\sin 	heta \approx 	heta$. This simplifies our equation to:

$$	heta = \frac{\lambda}{b}$$

The angular width of the central maximum is the total angle between the first minimum on the top and the first minimum on the bottom. Therefore, the angular width is $2	heta$, which equals $\frac{2\lambda}{b}$.

### The Proportionality Principle

From the formula $\frac{2\lambda}{b}$, we can clearly see that the angular width is directly proportional to the wavelength $\lambda$. The physical dimensions of the setup, namely the slit width $b$, remain constant.

The problem states that when a new wavelength is used, the angular width decreases by **30%**. Because of the direct proportionality, the new wavelength $\lambda'$ must also be 30% less than the original wavelength.

We can express this mathematically as:

$$\lambda' = \lambda - 0.30\lambda = 0.70\lambda$$

Given that the original wavelength $\lambda$ is $6000 	ext{ \AA}$, we can easily calculate the new wavelength:

$$\lambda' = 0.70 	imes 6000 	ext{ \AA} = 4200 	ext{ \AA}$$

This gives us the first part of our answer.

### The Immersion Effect

Now, let's explore what happens when the entire apparatus is immersed in a liquid. The problem tells us that this immersion causes the exact same 30% decrease in the angular width.

When light travels from air (or vacuum) into a denser medium like a liquid, its speed decreases. However, the frequency of the light remains constant because it is a property of the source.

Since wave speed is the product of frequency and wavelength ($v = f\lambda$), a decrease in speed must result in a proportional decrease in wavelength. The new wavelength in the medium is given by:

$$\lambda_{	ext{medium}} = \frac{\lambda_{	ext{air}}}{\mu}$$

Here, $\mu$ is the refractive index of the liquid.

### Final Calculation

We know that the wavelength in the liquid must be equal to the $4200 	ext{ \AA}$ we calculated earlier to produce the same 30% decrease in angular width.

We can set up the following equation using the original wavelength in air ($6000 	ext{ \AA}$):

$$4200 = \frac{6000}{\mu}$$

Solving for the refractive index $\mu$, we get:

$$\mu = \frac{6000}{4200} = \frac{10}{7}$$

Calculating the decimal value, we find:

$$\mu \approx 1.43$$

This gives us the refractive index of the liquid, completing our solution.

Visualized Solution

Visualizing the Setup

Angular Position of First Minimum