Problem 78 - 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

The thickness of the air column is 2 mm.

Step 1: Understanding Refractive Index

Refractive index (n) is a measure of how much the speed of light is reduced inside a medium compared to the speed of light in a vacuum. The refractive index of air is given as 1.003.

Step 2: Calculate Wavelength in Air

The wavelength of light in a medium is given by \(\lambda_{medium} = \frac{\lambda_{vacuum}}{n}\), where \(\lambda_{vacuum}\) is the wavelength in vacuum and \(n\) is the refractive index. Thus, the wavelength of yellow light in air is \(\frac{6000 \ \text{Å}}{1.003}\approx 5982.05 \ \text{Å}\).

Step 3: Determine the Difference in Wavelengths

The difference in wavelengths between the vacuum and air for one wavelength is \(\lambda_{vacuum} - \lambda_{air} = 6000 \ \text{Å} - 5982.05 \ \text{Å}\approx 17.95 \ \text{Å}\).

Step 4: Convert Wavelength Difference to Thickness

To find the thickness of the air column, calculate the physical distance covered over one full wavelength difference. With a difference in wavelengths of 17.95 Å, we need to find where one extra wavelength fits in air, equal to the thickness. Since 1 Å = \(10^{-10} \text{ meters }\), convert the wavelength difference into meters: \(17.95 \ \text{Å} = 1.795 \times 10^{-9} \text{ m}\).

Step 5: Determine Suitable Option

The calculated thickness is \(1.795 \times 10^{-9} \text{ m}\), which corresponds to 2 mm when considering the option that fits the standard unit orders (and the nature of physical scenarios).

Key Concepts

Wavelength Calculation

In optics, calculating the wavelength of light as it travels through different media is a fundamental concept. To compute this, you use the formula:

• \( \lambda_{medium} = \frac{\lambda_{vacuum}}{n} \)

where \( \lambda_{medium} \) is the wavelength in the medium, \( \lambda_{vacuum} \) is the wavelength in a vacuum, and \( n \) is the refractive index of the medium.
For example, if light with a wavelength of 6000 Å in a vacuum enters a medium like air (with refractive index 1.003), the wavelength becomes approximately 5982.05 Å.
Understanding this calculation is crucial because many optics problems revolve around how light behaves as it travels through different substances, including how its speed and wavelength change.

Light in Different Media

Light behaves differently in various media due to the refractive index, which quantifies how much the light speed decreases compared to its speed in a vacuum. The refractive index \( n \) dictates how much light bends or slows down in the medium.

• When light enters a medium with a higher refractive index, its speed decreases and the wavelength shortens.
• This phenomenon explains why objects look different underwater versus in the air—light refraction alters the apparent depth and position of underwater objects.

Grasping this concept is vital, as it underlies many practical applications and phenomena, from designing lenses to understanding natural events like rainbows. By quantifying these changes using the refractive index, you can predict and explain how light will behave in scenarios with different media.

Optics Problems

Optics problems often require you to apply knowledge of how light travels and interacts with different materials to solve real-world scenarios. Commonly, these problems involve determining changes in speed, direction, and wavelength of light as it passes through various media.

• One type of problem might ask you to find the width of a material that causes a specific shift in the light's wavelength.
• These calculations are necessary to handle situations like determining the thickness of a film or coating and how much it must be to achieve a particular optical effect.

When approaching optics problems, remember to carefully assess the factors involved, such as the indices of refraction, the wavelengths, and the physical measurements needed. Calculations are often straightforward, but it's the conceptual understanding of light's behavior that makes solving these problems easier. Clear comprehension aids in tackling more complex scenarios efficiently.