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 Question:
We are dealing with a Young's Double Slit Experiment (YDSE) using two different wavelengths. We need to find the physical distance separating the 3rd order bright fringes produced by each respective wavelength.

Step 2: Detailed Explanation:
The position ($y_n$) of the $n^{\text{th}}$ order bright fringe from the central maximum is given by the formula:
$y_n = \frac{n \lambda D}{d}$
where:
$n$ = fringe order = $3$
$D$ = screen distance = $1 \text{ m}$
$d$ = slit separation = $2 \text{ mm} = 2 \times 10^{-3} \text{ m}$
Position of 3rd bright fringe for $\lambda_1$:
$y_{3,1} = \frac{3 \lambda_1 D}{d}$
Position of 3rd bright fringe for $\lambda_2$:
$y_{3,2} = \frac{3 \lambda_2 D}{d}$
The separation ($\Delta y$) between these two fringes is:
$\Delta y = y_{3,2} - y_{3,1} = \frac{3D}{d} (\lambda_2 - \lambda_1)$
We are given the relationship $\lambda_2 = 1.5\lambda_1$.
Therefore, $(\lambda_2 - \lambda_1) = 1.5\lambda_1 - 1.0\lambda_1 = 0.5\lambda_1$.
Substitute this back into the separation equation:
$\Delta y = \frac{3D}{d} \times (0.5\lambda_1)$
Substitute the numerical values for $D$ and $d$:
$\Delta y = \frac{3(1)}{2 \times 10^{-3}} \times 0.5\lambda_1$
$\Delta y = \frac{1.5}{2 \times 10^{-3}} \lambda_1$
$\Delta y = 0.75 \times 10^3 \lambda_1$

Step 3: Final Answer:
The separation is $(0.75 \times 10^3)\lambda_1$, matching option (d).