Question:
When the distance between the slit and screen is doubled, the angular separation between two fringes in single-slit diffraction experiment will:
When the distance between the slit and screen is doubled, the angular separation between two fringes in single-slit diffraction experiment will:
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Fringe angles depend only on wavelength and slit width, not on the screen distance. The linear spacing doubles but the angle stays the same.
Updated On: Oct 21, 2026
- be doubled
- be halved
- remain constant
- become one-fourth
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
In single-slit diffraction, dark fringes (minima) form at angles \(\theta\) that satisfy \(a\sin\theta = n\lambda\). Here a is the slit width and \(\lambda\) is the wavelength.
The angle depends only on \(\lambda\) and a. The distance D to the screen does not appear in this condition.
Step 2: Key Formula or Approach:
For small angles, the angular position of the n-th minimum is \(\theta_n = n\lambda/a\). The angular separation of adjacent fringes is \(\lambda/a\).
The linear position on a screen at distance D is \(y_n = D\theta_n\), so the linear width changes with D, but the angle does not.
Step 3: Detailed Explanation:
When D is doubled, the fringes spread out: the linear separation \(y = D\lambda/a\) doubles.
The angle subtended at the slit is \(\theta = y/D = (2D\lambda/a)/(2D) = \lambda/a\). It is unchanged.
So the angular separation stays constant.
Step 4: Why the other options are wrong:
(A) Doubled: the linear width doubles, not the angular width. Angular separation does not depend on D.
(B) Halved: there is no reason for the angle to shrink when the screen moves away.
(D) One-fourth: this would need \(\theta \propto 1/D^2\), which is not true for diffraction.
Final Answer:
The angular separation depends only on \(\lambda/a\), so it remains constant. This is option (C). \[ \boxed{\text{remains constant}} \]
In single-slit diffraction, dark fringes (minima) form at angles \(\theta\) that satisfy \(a\sin\theta = n\lambda\). Here a is the slit width and \(\lambda\) is the wavelength.
The angle depends only on \(\lambda\) and a. The distance D to the screen does not appear in this condition.
Step 2: Key Formula or Approach:
For small angles, the angular position of the n-th minimum is \(\theta_n = n\lambda/a\). The angular separation of adjacent fringes is \(\lambda/a\).
The linear position on a screen at distance D is \(y_n = D\theta_n\), so the linear width changes with D, but the angle does not.
Step 3: Detailed Explanation:
When D is doubled, the fringes spread out: the linear separation \(y = D\lambda/a\) doubles.
The angle subtended at the slit is \(\theta = y/D = (2D\lambda/a)/(2D) = \lambda/a\). It is unchanged.
So the angular separation stays constant.
Step 4: Why the other options are wrong:
(A) Doubled: the linear width doubles, not the angular width. Angular separation does not depend on D.
(B) Halved: there is no reason for the angle to shrink when the screen moves away.
(D) One-fourth: this would need \(\theta \propto 1/D^2\), which is not true for diffraction.
Final Answer:
The angular separation depends only on \(\lambda/a\), so it remains constant. This is option (C). \[ \boxed{\text{remains constant}} \]
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