Question

In Young's double slit experiment, the distance between the slits is 2 mm and the slits are 1 m away from the screen. Two interference patterns can be obtained on the screen due to light of wavelength '$\lambda_1$' and '$\lambda_2$' respectively. The separation on the screen between the 3rd order bright fringes on the two interference patterns is ($\lambda_2 = 1.5\lambda_1$) ______.

• A. $(0.75 \times 10^{-3})\lambda_1$
• B. $(1.75 \times 10^{-3})\lambda_1$
• C. $(2.00 \times 10^{-3})\lambda_1$
• D. $(0.75 \times 10^3)\lambda_1$

Hint:

Be extremely careful with powers of 10! The slit distance $d$ is in the denominator, so a $10^{-3}$ on the bottom flips up to become a $10^{+3}$ on the top!

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The position of the $n$-th bright fringe is given by $y_n = \frac{n \lambda D}{d}$.

Step 2: Formula Application:
$n = 3$, $D = 1$ m, $d = 2$ mm $= 2 \times 10^{-3}$ m.
$\lambda_2 = 1.5\lambda_1$.

Step 3: Explanation:
$y_{3(\lambda_1)} = \frac{3 \lambda_1 (1)}{2 \times 10^{-3}}$.
$y_{3(\lambda_2)} = \frac{3 (1.5 \lambda_1) (1)}{2 \times 10^{-3}}$.
Separation $\Delta y = y_{3(\lambda_2)} - y_{3(\lambda_1)} = \frac{3 \lambda_1 (1.5 - 1)}{2 \times 10^{-3}}$.
$\Delta y = \frac{3 \lambda_1 (0.5)}{2 \times 10^{-3}} = \frac{1.5 \lambda_1}{2 \times 10^{-3}} = 0.75 \times 10^3 \lambda_1$ (Wait, re-checking exponent).
The $10^{-3}$ in the denominator becomes $10^3$ in the numerator. However, wavelengths are usually $\sim 10^{-7}$, making the result look small.
$\Delta y = \frac{0.75 \lambda_1}{10^{-3}} = 0.75 \times 10^3 \lambda_1$.

Step 4: Final Answer:
The separation is $(0.75 \times 10^3)\lambda_1$. (Note: Option A usually implies $10^3$ in these formats).