Question

In Young’s double-slit experiment, the wavelength of light is doubled while all other parameters remain unchanged. The fringe width becomes:

• A. Half of the original
• B. Unchanged
• C. Double the original
• D. Four times the original

Hint:

Since $\beta$ is directly proportional to $\lambda$, any modification factor applied to the wavelength is directly transferred to the fringe width.
If $\lambda$ is doubled, $\beta$ is doubled.
If $\lambda$ is halved, $\beta$ is halved.
This direct proportionality simplifies conceptual questions in YDSE.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question is from "Wave Optics" and concerns the dependence of the fringe width ($\beta$) in Young's Double Slit Experiment (YDSE) on the wavelength of light ($\lambda$) when all other experimental parameters are held constant.

Step 2: Key Formula or Approach:
The fringe width ($\beta$) in a standard YDSE is defined by:
\[ \beta = \frac{\lambda D}{d} \]
where $\lambda$ is the wavelength, $D$ is the distance to the screen, and $d$ is the slit separation.

Step 3: Detailed Explanation:

• From the formula for fringe width, we can see that:
\[ \beta \propto \lambda \]
This means the fringe width is directly proportional to the wavelength of the light used, provided the screen distance ($D$) and the slit separation ($d$) remain unchanged.

• Let the initial wavelength be $\lambda_1 = \lambda$, and the corresponding initial fringe width be $\beta_1 = \beta$.

• The new wavelength is doubled: $\lambda_2 = 2\lambda$.

• The new fringe width $\beta_2$ is:
\[ \beta_2 = \frac{\lambda_2 D}{d} = \frac{2\lambda D}{d} = 2\beta_1 \]

• Thus, doubling the wavelength results in exactly doubling the fringe width.



Step 4: Final Answer:
The fringe width becomes double the original, which corresponds to option (C).