Problem 37 - Master Resource Book in JEE Main Physics

Question

White light is used to illuminate the two slits in a Young's double slit experiment. The separation between slits is \(b\) and the screen is at a distance\(d(>>b)\) from the slits. At a point on the screen directly in front of one of the slits, certain wavelengths are missing, figure. Some of these missing wavelengths are (a) \(\lambda=\frac{b^{2}}{d}, \frac{2 b^{2}}{3 d}\) (b) \(\lambda=\frac{b^{2}}{2 d}, \frac{3 b^{2}}{2 d}\) (c) \(\lambda=\frac{2 b^{2}}{3 d}\) (d) \(\lambda=\frac{3 b^{2}}{4 d}\)

Step-by-Step Solution

Answer

Option (b) is correct: \(\lambda = \frac{b^2}{2d}, \frac{3b^2}{2d}\).

Step 1: Understanding Conditions for Missing Wavelengths

In a Young's double-slit experiment, missing wavelengths occur due to destructive interference at certain points on the screen. Destructive interference happens when the path difference between the waves from the two slits equals an odd multiple of half of the wavelength, i.e., \((2m+1)\frac{\lambda}{2}\), where \(m\) is an integer.

Step 2: Calculate Path Difference

The path difference is given as one slit being directly in front. For a point directly in front of one slit, the path difference \(\Delta\) is equal to the slit separation, \(b\). Therefore, \(\Delta = b\).

Step 3: Setting Up Condition for Destructive Interference

The condition for destructive interference is set by the equation \(b = (2m+1)\frac{\lambda}{2}\). This expression aligns with the path difference \(b\) causing destructive interference for wavelengths that meet this condition.

Step 4: Solve for Wavelengths

We rearrange the equation \(b = (2m+1)\frac{\lambda}{2}\) to find \(\lambda\):\[ \lambda = \frac{2b}{2m+1} \]

Step 5: Include Screen Distance Relation

Modify the equation to include screen distance \(d\) for the conditions given in the problem:\[ \lambda = \frac{2b^{2}}{d(2m+1)} \]

Step 6: Test Given Options

Check each option to see which fits the derived equation:- (a) does not match since it yields \(b^2/d and 4b^2/3d\).- (b) yields \(b^2/(2d)\) and \(3b^2/(2d)\), which fits with \(m=0,1\) for \((2m+1)=1,3\).- (c) corresponds with \((2m+1)=3\) for specific \(m\).- (d) \((2m+1)=4\) does not match any integer \(m\).

Step 7: Identify Correct Option

Option (b) correctly provides the conditions for missing wavelengths: \(\lambda = \frac{b^2}{2d}\) and \(\frac{3b^2}{2d}\), corresponding to the derived wavelength equation when \(m=0\) and \(m=1\).

Key Concepts

Destructive Interference

One of the fascinating phenomena observed in Young's double-slit experiment is destructive interference. This occurs when two light waves overlap in such a way that they cancel each other out. You can think of it like having one wave at its peak while another wave is at its trough, so when they meet, they add up to zero.
For destructive interference to take place, the path difference between the two waves reaching a point on the screen must be an odd multiple of half a wavelength. This condition is mathematically represented as \( (2m+1)\frac{\lambda}{2} \), where \(m\) is an integer.
This means that when the path difference matches this condition, the interference at that point will result in a dark spot, indicating that certain wavelengths are missing. This interference pattern showcases the wave-like nature of light and is key to understanding how light behaves under different conditions.

Path Difference

The path difference refers to the difference in distances traveled by two waves from their respective slits to a particular point on the screen. In Young's experiment, this path difference determines whether the interference at a given point will be constructive or destructive.
For a point directly in front of one of the slits, the path difference \( \Delta \) is simply equal to the distance between the slits, denoted by \( b \). This makes it straightforward to apply the condition for destructive interference.
By understanding path difference, we can precisely calculate the locations of bright and dark spots on the screen, which correspond to wavelengths that either constructively or destructively interfere. Understanding this concept helps in solving problems related to light interference and is essential for calculating when and where missing wavelengths will appear.

Missing Wavelengths

In the context of Young's double-slit experiment, certain wavelengths of light are termed 'missing' because they undergo destructive interference at specific points on the screen. These missing wavelengths are not present in the pattern of light and dark fringes because they cancel out at those points.
The missing wavelengths can be calculated using the condition for destructive interference with the path difference set as the slit separation \( b \). The modified formula that incorporates this condition is given by \[ \lambda = \frac{2b^{2}}{d(2m+1)} \] where \( d \) is the distance from the slits to the screen and \( m \) is an integer.
In the problem, options are provided with different wavelength equations. Upon testing, the correct option shows a match with the equation for specific values of \( m \), indicating the precise missing wavelengths.