Question:
The maximum intensity in a Young's double slit experiment is \( I_o \). Distance between the slits (\( d \)) is \( 5\lambda \), where \( \lambda \) is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at \( D = 10d \) is \dots
The maximum intensity in a Young's double slit experiment is \( I_o \). Distance between the slits (\( d \)) is \( 5\lambda \), where \( \lambda \) is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at \( D = 10d \) is \dots
Updated On: Apr 29, 2026
- \( I_o/4 \)
- \( I_o/2 \)
- \( I_o \)
- \( 3I_o/4 \)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Intensity at a point on the screen depends on the phase difference between the two waves. The phase difference is determined by the path difference from the slits to that point.
: Key Formula or Approach:
Path difference: \( \Delta x = \frac{dy}{D} \).
Phase difference: \( \phi = \frac{2\pi}{\lambda} \Delta x \).
Resultant Intensity: \( I = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \).
Step 2: Detailed Explanation:
The point "exactly opposite to one of the slits" is at a height \( y = d/2 \) from the central axis.
Given: \( d = 5\lambda \), \( D = 10d = 50\lambda \), and \( y = d/2 = 2.5\lambda \).
Calculate path difference \( \Delta x \):
\[ \Delta x = \frac{d \cdot y}{D} = \frac{d \cdot (d/2)}{10d} = \frac{d}{20} \]
Substitute \( d = 5\lambda \):
\[ \Delta x = \frac{5\lambda}{20} = \frac{\lambda}{4} \]
Calculate phase difference \( \phi \):
\[ \phi = \frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2} \]
Calculate intensity \( I \):
\[ I = I_o \cos^2\left(\frac{\phi}{2}\right) = I_o \cos^2\left(\frac{\pi/2}{2}\right) = I_o \cos^2\left(\frac{\pi}{4}\right) \]
\[ I = I_o \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_o}{2} \]
Step 3: Final Answer:
The intensity of the fringe is \( I_o/2 \).
Intensity at a point on the screen depends on the phase difference between the two waves. The phase difference is determined by the path difference from the slits to that point.
: Key Formula or Approach:
Path difference: \( \Delta x = \frac{dy}{D} \).
Phase difference: \( \phi = \frac{2\pi}{\lambda} \Delta x \).
Resultant Intensity: \( I = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \).
Step 2: Detailed Explanation:
The point "exactly opposite to one of the slits" is at a height \( y = d/2 \) from the central axis.
Given: \( d = 5\lambda \), \( D = 10d = 50\lambda \), and \( y = d/2 = 2.5\lambda \).
Calculate path difference \( \Delta x \):
\[ \Delta x = \frac{d \cdot y}{D} = \frac{d \cdot (d/2)}{10d} = \frac{d}{20} \]
Substitute \( d = 5\lambda \):
\[ \Delta x = \frac{5\lambda}{20} = \frac{\lambda}{4} \]
Calculate phase difference \( \phi \):
\[ \phi = \frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2} \]
Calculate intensity \( I \):
\[ I = I_o \cos^2\left(\frac{\phi}{2}\right) = I_o \cos^2\left(\frac{\pi/2}{2}\right) = I_o \cos^2\left(\frac{\pi}{4}\right) \]
\[ I = I_o \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_o}{2} \]
Step 3: Final Answer:
The intensity of the fringe is \( I_o/2 \).
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