In `YDSE` ratio of width of slit is `4:1`, then ratio of maximum to minimum intensity
In `YDSE` ratio of width of slit is `4:1`, then ratio of maximum to minimum intensity
A
`25:9`
B
`9:1`
C
`4:1`
D
`(sqrt(3)-1)^(4):16`
Text Solution
AI Generated Solution
The correct Answer is:
To solve the problem of finding the ratio of maximum to minimum intensity in Young's double slit experiment (YDSE) when the ratio of the widths of the slits is 4:1, we can follow these steps:
### Step-by-Step Solution:
1. **Understand the relationship between intensity and slit width**:
In YDSE, the intensity of light from each slit is proportional to the square of the amplitude of the light wave, which in turn is proportional to the width of the slit. Therefore, if we denote the widths of the two slits as \( W_1 \) and \( W_2 \), we can express the intensities \( I_1 \) and \( I_2 \) as:
\[
I_1 \propto W_1 \quad \text{and} \quad I_2 \propto W_2
\]
2. **Assign values based on the given ratio**:
Given the ratio of the widths of the slits is \( 4:1 \), we can assign:
\[
W_1 = 4k \quad \text{and} \quad W_2 = k
\]
where \( k \) is a constant.
3. **Calculate the intensities**:
From the relationship between intensity and width, we have:
\[
I_1 = k_1 W_1 = k_1 (4k) = 4k_1 k
\]
\[
I_2 = k_2 W_2 = k_2 (k) = k_2 k
\]
For simplicity, we can assume \( k_1 = k_2 = 1 \) (the proportionality constants can be ignored for the ratio), thus:
\[
I_1 = 4k \quad \text{and} \quad I_2 = k
\]
4. **Calculate the maximum and minimum intensities**:
The maximum intensity \( I_{\text{max}} \) and minimum intensity \( I_{\text{min}} \) can be calculated using the formulas:
\[
I_{\text{max}} = (\sqrt{I_1} + \sqrt{I_2})^2
\]
\[
I_{\text{min}} = (\sqrt{I_1} - \sqrt{I_2})^2
\]
Substituting the values:
\[
I_{\text{max}} = (\sqrt{4k} + \sqrt{k})^2 = (2\sqrt{k} + \sqrt{k})^2 = (3\sqrt{k})^2 = 9k
\]
\[
I_{\text{min}} = (\sqrt{4k} - \sqrt{k})^2 = (2\sqrt{k} - \sqrt{k})^2 = (\sqrt{k})^2 = k
\]
5. **Find the ratio of maximum to minimum intensity**:
Now, we can find the ratio:
\[
\frac{I_{\text{max}}}{I_{\text{min}}} = \frac{9k}{k} = 9
\]
### Final Answer:
The ratio of maximum to minimum intensity is \( 9:1 \).
Topper's Solved these Questions
Similar Questions
Explore conceptually related problems
If the ratio of amplitude of two waves is 4:3 , then the ratio of maximum and minimum intensity is
The width of one of the two slits in a Young's double slit experiment is double of the other slit. Assuming that the amplitude of the light coming from a slit is proportional to the slit width, find the ratio of the maximum to the minimum intensity in the interference pattern.
The width of one of the two slits in a Young's double slit experiment is double of the other slit. Assuming that the amplitude of the light coming from a slit is proportional to the slit width, find the ratio of the maximum to the minimum intensity in the interference pattern.
Two waves having the intensities in the ratio of 9 : 1 produce interference. The ratio of maximum to minimum intensity is equal to
If two light waves having same frequency have intensity ratio 4:1 and they interfere, the ratio of maximum to minimum intensity in the pattern will be
A transparent glass plate of thickness 0.5 mm and refractive index 1.5 is placed infront of one of the slits in a double slit experiment. If the wavelength of light used is 6000 A^(@) , the ratio of maximum to minimum intensity in the interference pattern is 25/4. Then the ratio of light intensity transmitted to incident on thin transparent glass plate is
The intensity of the light coming from one of the slits in a Young's double slit experiment is double the intensity from the other slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed.
The ratio of the intensities of two interfering waves is 81:1. What is the ratio of the maximum to minimum intensity?
In Young's double slit experiment, the ratio of maximum and minimum intensities in the fringe system is 9:1 the ratio of amplitudes of coherent sources is
The ratio of intensities of two waves is 9 : 1 When they superimpose, the ratio of maximum to minimum intensity will become :-