Question

In a Young's double-slit experiment, fringes are obtained on a screen placed at a distance $D$ from the slits. It is observed that if the screen is moved $5\,\text{cm}$ towards the slits, the fringe width changes by $0.03\,\text{mm}$. If the distance between the slits is $1\,\text{mm}$, calculate the wavelength of the light used.

Hint:

Use \(\Delta\beta = \lambda\,\Delta D/d\), since \(\beta = \lambda D/d\).

Answer 1

Step 1: Understanding the Concept:
In Young's experiment the fringe width is \(\beta = \dfrac{\lambda D}{d}\). It is proportional to the screen distance D.
Moving the screen closer to the slits makes the fringes narrower.

Step 2: Key Formula or Approach:
Change in fringe width \(\Delta\beta = \dfrac{\lambda\,\Delta D}{d}\). So \(\lambda = \dfrac{d\,\Delta\beta}{\Delta D}\).

Step 3: Substitute the data:
\(d = 1\ \text{mm} = 1 \times 10^{-3}\ \text{m}\), \(\Delta\beta = 0.03\ \text{mm} = 3 \times 10^{-5}\ \text{m}\), \(\Delta D = 5\ \text{cm} = 5 \times 10^{-2}\ \text{m}\).
\[ \lambda = \frac{(1 \times 10^{-3})(3 \times 10^{-5})}{5 \times 10^{-2}} \]
\[ \lambda = 6 \times 10^{-7}\ \text{m} = 600\ \text{nm} \]

Final Answer:
The wavelength of the light is 600 nm. \[ \boxed{\lambda = 6 \times 10^{-7}\ \text{m}} \]