Question

The distance between the first and sixth minima in the diffraction pattern of a single slit, it is $0.5\mathrm{mm}$. The screen is $0.5\mathrm{m}$ away from the Slit. If the wavelength of light is $5000\text{AA}$, then the width of the slit will be $\mathrm{mm}$ (D) $1.0$ (A) 5 (B) $2.5$ (C) $1.25$

Short answer

The width of the slit is (C) $1.25\mathrm{mm}$.

Step-by-step solution

Understand the formula for minima in a single-slit diffraction pattern.

The formula for minima in a single-slit diffraction pattern is given by: $$m\lambda =a\sin \theta$$. Here, $m$ is the order of the minima, $\lambda$ is the wavelength of light, $a$ is the width of the slit, and $\theta$ is the angle formed between the central maximum and the minima. We need to find $a$.

Convert the given measurements to meters.

We have been given the distance between the slit and the screen $L=0.5\text{m}$, the wavelength of light $\lambda =5000\text{AA}$, and the distance between the first and sixth minima $y=0.5\text{mm}$. Let's convert them to meters: $\lambda =5000\text{AA}=5\times {10}^{-7}\text{m}$ $y=0.5\text{mm}=5\times {10}^{-4}\text{m}$

Calculate the angle θ using the distance between the first and sixth minima.

First, we need to find the distance between consecutive minima: $\mathrm{\Delta }y=\frac{5\times {10}^{-4}\text{m}}{5}=1\times {10}^{-4}\text{m}$. Next, since the angle $\theta$ is small, we can use the approximation $\tan \theta \approx \sin \theta \approx \frac{\mathrm{\Delta }y}{L}\Rightarrow \sin \theta =\frac{1\times {10}^{-4}\text{m}}{0.5\text{m}}=2\times {10}^{-4}$.

Use the formula to find the width of the slit and select the correct answer.

Now, we can use the formula for minima with the given values: $m\lambda =a\sin \theta \Rightarrow a=\frac{m\lambda }{\sin \theta }$ We are given that the distance between the first and sixth minima is 0.5 mm, which means we need to consider the distance for 5 orders (from first to the sixth minimum is a difference of 5 orders). Thus, $m=5$. Now, substitute the values we found: $a=\frac{5(5\times {10}^{-7}\text{m})}{2\times {10}^{-4}}=\frac{25\times {10}^{-7}\text{m}}{2\times {10}^{-4}}=1.25\times {10}^{-3}\text{m}$ Converting the slit width back to millimeters, we get, $a=1.25\text{mm}$. Therefore, the correct answer is (C) $1.25$.

Key concepts

Minima in Diffraction

In single-slit diffraction patterns, **minima** are the dark spots that appear where the light waves interfere destructively. To determine the position of these minima, we use the formula: $$m\lambda =a\sin \theta$$, where:

• $m$ is the order of the minima (e.g., first, second, etc.).
• $\lambda$ is the wavelength of the light used.
• $a$ stands for the width of the slit, which the light passes through.
• $\theta$ is the angle of diffraction.

When analyzing a diffraction pattern, the order of a minimum helps define how far that specific dark band is from the central bright spot. For instance, in the problem, the minima from the first to the sixth are considered, indicating 5 intervals of minima.

Wavelength Conversion

**Wavelength conversion** is essential to ensure all measurements are consistent before performing calculations. Commonly, wavelengths can be given in Ångströms ($\AA$), millimeters, or meters. In physics, SI units are preferable for accuracy. Hence, it's crucial to convert to meters.

Converting the wavelength from Ångströms to meters involves:

• Since 1 Ångström equals ${10}^{-10}$ meters, for instance, $5000$ Å becomes $5\times {10}^{-7}$ meters.

Converting measurements allows for seamless computation and accuracy when implementing them into the diffraction formulas. Consistent unit usage helps avoid errors in calculating phenomena like diffraction patterns.

Diffraction Angle Calculation

When light passes through a single slit, **diffraction angles** can be calculated based on geometric approximations. Knowing how angles form in these patterns helps determine where minima appear.

For small angles, we can approximate:

• $\tan \theta \approx \sin \theta \approx \frac{y}{L}$.

Where:

• $y$ is the distance between the minima.
• $L$ is the distance from the slit to the screen where the diffraction pattern is projected.

By using this approximation, it simplifies the complexity that could arise from trigonometric computations.For example, in the problem discussed, calculating $\sin \theta$ provided a clearer path to find the slit width without unnecessary complications.

Slit Width Determination

**Slit width determination** is crucial in understanding how light diffracts when it meets an obstacle. The diffraction formula allows us to solve for the slit width $a$ once we know the minima's order $m$, wavelength, and calculated angle.

The calculation involves rearranging:

• $$a=\frac{m\lambda }{\sin \theta }$$

By substituting known variables, the formula tells us how wide the slit is in millimeters or meters. In this exercise, converting units back to millimeters after calculating in meters was necessary for final clarity and alignment with given choices.

This approach is highly useful in both experimental physics and theoretical studies where determining geometric parameters of a slit can predict diffraction behaviors.