Question

In Young's double slit experiment, the $10^{\text{th}}$ maximum of wavelength '$\lambda_1$' is at a distance of '$Y_1$' from the central maximum. When the wavelength of the source is changed to '$\lambda_2$', the $5^{\text{th}}$ maximum is at a distance '$Y_2$' from the central maximum. The ratio $\frac{Y_1}{Y_2}$ is

• A. $\frac{2\lambda_1}{\lambda_2}$
• B. $\frac{\lambda_2}{2\lambda_1}$
• C. $\frac{2\lambda_2}{\lambda_1}$
• D. $\frac{\lambda_1}{2\lambda_2}$

Hint:

You can solve this very quickly by setting up a simple scale balance: $Y_1 / Y_2 = (n_1 / n_2) \times (\lambda_1 / \lambda_2)$. Since the first order is 10 and the second order is 5, the order ratio is exactly $10 / 5 = 2$. Multiplying this by the wavelength ratio gives $\frac{2\lambda_1}{\lambda_2}$ in a single thought step!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
Relate the fringe widths (Y₁, Y₂) in a double-slit interference pattern to the diffraction orders (n₁, n₂) and wavelengths (λ₁, λ₂).

Step 2: Key Formula or Approach:
The fringe position scales proportionally with both the order number and the wavelength: Y ∝ n × λ. Thus, the ratio Y₁/Y₂ = (n₁/n₂) × (λ₁/λ₂).

Step 3: Detailed Explanation:
Given the first order n₁ = 10 and second order n₂ = 5, the order ratio is n₁/n₂ = 10/5 = 2. The fringe width ratio therefore becomes Y₁/Y₂ = 2 × (λ₁/λ₂) = 2λ₁/λ₂. This direct proportional reasoning bypasses lengthy derivations and gives the answer in a single mental step.

Step 4: Final Answer:
The ratio simplifies directly to 2λ₁/λ₂.