Question:
In a biprism experiment, a steady interference pattern is observed on the screen kept at a distance of 100 cm using a light of wavelength \(5000\) Å. Without changing the distance between the virtual images of the slit, the source of light is replaced by a source of wavelength \(6400\) Å. Now, to reduce the fringe width by \(20\%\) of its initial value, the screen should be moved
In a biprism experiment, a steady interference pattern is observed on the screen kept at a distance of 100 cm using a light of wavelength \(5000\) Å. Without changing the distance between the virtual images of the slit, the source of light is replaced by a source of wavelength \(6400\) Å. Now, to reduce the fringe width by \(20\%\) of its initial value, the screen should be moved
Show Hint
Fringe width is proportional to lambda times D, so set the new product to 80 percent of the old one.
Updated On: Oct 1, 2026
- towards the source by \(37.5\) cm
- towards the source by \(62.5\) cm
- away from the source by \(62.5\) cm
- away from the source by \(37.5\) cm
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The Correct Option is A
Solution and Explanation
Step 1: Understand the concept
In a biprism experiment, the fringe width is \(\beta = \dfrac{\lambda D}{d}\). Here \(d\), the separation between the virtual images, is unchanged.
Step 2: Set up the condition
Initially \(\beta_1 = \dfrac{5000\times100}{d}\) (taking \(D\) in cm and \(\lambda\) in Angstrom, the units cancel in the ratio). After changing the source and the screen, we want \(\beta_2 = 0.8\,\beta_1\):
\[ 6400\,D' = 0.8\times5000\times100 \]
Step 3: Solve for the new distance
\[ D' = \frac{400000}{6400} = 62.5\ \text{cm} \]
Step 4: Result
The screen goes from 100 cm to 62.5 cm, so it is moved by \(37.5\) cm towards the source (the biprism), option (A). The other options: 62.5 cm is the new distance, not the movement, and moving away would increase the fringe width.
Final Answer:
The screen moves 37.5 cm towards the source. This is option (A). \[ \boxed{\text{(A) }\text{towards the source by 37.5 cm}} \]
In a biprism experiment, the fringe width is \(\beta = \dfrac{\lambda D}{d}\). Here \(d\), the separation between the virtual images, is unchanged.
Step 2: Set up the condition
Initially \(\beta_1 = \dfrac{5000\times100}{d}\) (taking \(D\) in cm and \(\lambda\) in Angstrom, the units cancel in the ratio). After changing the source and the screen, we want \(\beta_2 = 0.8\,\beta_1\):
\[ 6400\,D' = 0.8\times5000\times100 \]
Step 3: Solve for the new distance
\[ D' = \frac{400000}{6400} = 62.5\ \text{cm} \]
Step 4: Result
The screen goes from 100 cm to 62.5 cm, so it is moved by \(37.5\) cm towards the source (the biprism), option (A). The other options: 62.5 cm is the new distance, not the movement, and moving away would increase the fringe width.
Final Answer:
The screen moves 37.5 cm towards the source. This is option (A). \[ \boxed{\text{(A) }\text{towards the source by 37.5 cm}} \]
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