Problem 77 - Master Resource Book in JEE Main Physics

Question

Air has refractive index \(1.003 .\) The thickness of air column, which will have one more wavelength of yellow light \((6000 \AA)\) than in the same thickness of vacuum is (a) \(2 \mathrm{~mm}\) (b) \(2 \mathrm{~cm}\) (c) \(2 \mathrm{~m}\) (d) \(2 \mathrm{~km}\)

Step-by-Step Solution

Answer

(a) 2 mm

Step 1: Understand the Problem

We are asked to find the thickness of an air column that contains one more wavelength of yellow light \((6000 \text{ Å})\) compared to the same thickness of vacuum. The refractive index of air is given as \(1.003\).

Step 2: Define the Relevant Formulas

The number of wavelengths \(n\) in a medium is given by \(n = \frac{t}{\lambda/n_i}\), where \(t\) is the thickness of the medium, \(\lambda\) is the wavelength in vacuum, and \(n_i\) is the refractive index. In air, the wavelength becomes \(\lambda/n_i\).

Step 3: Compute the Wavelengths Count Difference

We need one additional wavelength in air than in vacuum. Therefore:\[\text{Number of wavelengths in air} = \frac{t}{\lambda/1.003} = t \cdot \frac{1.003}{\lambda}\]\[\text{Number of wavelengths in vacuum} = \frac{t}{\lambda}\]Thus, the increment of one wavelength is expressed as:\[t \cdot \frac{1.003}{\lambda} = \frac{t}{\lambda} + 1\]

Step 4: Solve for Thickness \(t\)

Re-arrange the increment equation to find \(t\):\[t \cdot \left( \frac{1.003}{\lambda} - \frac{1}{\lambda} \right) = 1\]\[t \cdot \frac{0.003}{\lambda} = 1\]\[t = \frac{\lambda}{0.003}\]Substitute \(\lambda = 6000 \text{ Å} = 6000 \times 10^{-10} \text{ m}\) into the equation:\[t = \frac{6000 \times 10^{-10}}{0.003}\]Calculate to find \(t\):\[t = 2 \times 10^{-3} \text{ m}\]

Step 5: Convert and Choose Correct Option

Convert the thickness \(t = 2 \times 10^{-3} \text{ m}\) to centimeters:\[2 \times 10^{-3} \text{ m} = 2 \text{ mm}\]Compare this to the options given:(a) \(2 \text{ mm}\)(b) \(2 \text{ cm}\)(c) \(2 \text{ m}\)(d) \(2 \text{ km}\)Hence, the correct answer is (a) \(2 \text{ mm}\).

Key Concepts

Wavelength Calculation

Wavelength calculation is a fundamental step in understanding optical physics concepts. Wavelength, denoted by \( \lambda \), is the distance between two consecutive peaks or troughs in a wave. This can change based on the medium through which light travels. To calculate the number of wavelengths in a given medium, we must consider the medium's refractive index.The formula \( n = \frac{t}{\lambda/n_i} \) helps in determining the number of wavelengths that fit into a specific thickness \( t \) of a medium. Here, \( \lambda \) represents the wavelength in a vacuum, while \( n_i \) represents the refractive index of the medium. By using this formula, we account for how light bends and slows down in different materials.Fundamentally, more wavelengths can fit into a medium with a higher refractive index than the same thickness in a vacuum, which is vital in optical physics.

Medium Thickness

The thickness of a medium can influence the number of wavelengths it accommodates. In this context, the thickness \( t \) of an air column needed to contain one additional wavelength compared to a vacuum is our focus. Using refractive index values, the relationship between wavelength and medium is expressed in terms of thickness. Intuitively, a thin medium will accommodate fewer waves, whereas a thicker medium can contain more waves of light.To determine the precise thickness \( t \), we solve the equation: \[ t \cdot \left( \frac{1.003}{\lambda} - \frac{1}{\lambda} \right) = 1 \]Simplified further to \( t = \frac{\lambda}{0.003} \), this calculation shows how to adjust medium thickness to achieve a specific optical effect, such as adding one more wavelength.

Optical Physics

Optical physics explores how light interacts with various materials. This exercise exemplifies a core principle of optical physics: the role of refractive index in altering light behavior. The refractive index of a material indicates how much it reduces light speed and bends light rays. For instance, a refractive index of 1.003 for air is close to that of a vacuum, indicating slight changes in light propagation. Understanding the effect of refractive index is crucial in designing lenses and optical instruments, as it helps predict how light will move through them. This involves calculating how light transitions from one medium to another, altering its wavelength and speed. In practical terms, optical physics aids in designing better glasses, cameras, and even understanding atmospheric optics, showing how such calculations and knowledge directly impact technology and daily life.