Question

In a Young's double-slit experiment with monochromatic light of wavelength 600 nm, and slit separation of 2 mm, fringes are observed on a screen at some distance. Calculate the change in fringe width if the screen is moved towards the slits by 10 cm.

Hint:

Use beta = lambda D / d. Only D changes, so change in beta = lambda x (change in D) / d.

Answer 1

Step 1: Write the formula:
The fringe width is \(\beta = \dfrac{\lambda D}{d}\). Here \(\lambda\) and \(d\) do not change; only \(D\) changes.
So the change in fringe width is \(\Delta\beta = \dfrac{\lambda\, \Delta D}{d}\).

Step 2: Put in the numbers:
\(\lambda = 600 \text{ nm} = 6 \times 10^{-7}\) m, \(d = 2 \text{ mm} = 2 \times 10^{-3}\) m, \(\Delta D = 10 \text{ cm} = 0.1\) m.
\[ \Delta\beta = \frac{6 \times 10^{-7} \times 0.1}{2 \times 10^{-3}} \]
\[ \Delta\beta = \frac{6 \times 10^{-8}}{2 \times 10^{-3}} = 3 \times 10^{-5} \text{ m} \]

Step 3: Say whether it rises or falls:
The screen moves towards the slits, so \(D\) becomes smaller. The fringe width therefore decreases.

Final Answer:
The fringe width decreases by \(3 \times 10^{-5}\) m, which is 0.03 mm. \[ \boxed{\Delta\beta = 3 \times 10^{-5} \text{ m} = 0.03 \text{ mm (decrease)}} \]